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Digital Image Processing 3
I.Introduction Spatial Domain Process
( , ) [ ( , )])
( , ) : input image
( , ) :output image
: an operator on defined over
a neighborhood of point ( , )
g x y T f x y
f x y
g x y
T f
x y
=
Digital Image Processing
4
I.Introduction
Spatial Domain Process
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Digital Image Processing 5
II.IntensitytransformationfunctionIntensity transformation function
( )s T r=
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6
II.Intensitytransformationfunction
Some basic intensity transformation
Functions
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Digital Image Processing 7
II.1.NegativeImage negatives
1s L r=
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8
II.1.Negative
Smalllesion
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Digital Image Processing 9
II.2.LogTransformLog Transformations
log(1 )s c r= +
Digital Image Processing
10
II.2.LogTransform
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Digital Image Processing 11
II.3.Power Law
s cr=
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12
II.3.Power Law
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Digital Image Processing 13
II.3.Power Law
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14
II.4.Piecewise-LinearTransform..
Contrast Stretching Expands the range of intensity levels in an image so
that it spans the full intensity range of the recording
medium or display device
Intensity-level Slicing Highlighting a specific range of intensities in an image
often is of interest
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Digital Image Processing 15
II.4.Piecewise-LinearTransform
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16
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Digital Image Processing 17
II.5.Bit PlaneSlicing
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18
II.5.Bit PlaneSlicing
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Digital Image Processing 19
III.Histogramprocessing Histogram Equalization
Histogram Matching
Local Histogram Processing
Using Histogram Statistics for Image
Enhancement
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20
III.Histogramprocessing
Histogram ( )
is the intensity value
is the number of pixels in the image with intensity
k k
th
k
k k
h r n
r k
n r
=
Normalized histogram ( )
: the number of pixels in the image of
size M N with intensity
kk
k
k
np r
MN
n
r
=
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Digital Image Processing 21
III.Histogramprocessing
4x4 image
Gray scale = [0,9]histogram
0 1
1
2
2
3
3
4
4
5
5
6
6
7 8 9
No. of pixels
Gray level
2 3 3 2
4 2 4 3
3 2 3 5
2 4 2 4
Digital Image Processing
22
III.Histogramprocessing
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Digital Image Processing 23
III.1.HistogramEqualization As the low-contrast images histogram is narrow
and centered toward the middle of the gray scale,
if we distribute the histogram to a wider range the
quality of the image will be improved.
We can do it by adjusting the probability density
function of the original histogram of the image so
that the probability spread equally
Digital Image Processing
24
III.1.HistogramEqualization
Histogram transformation
0 1rk
sk= T(rk)
s
r
T(r)
s = T(r)s = T(r)
Where 0 r 1
T(r) satisfies (a). T(r) is single-valued
and monotonicallyincreasingly in the interval0 r 1
(b). 0 T(r) 1 for0 r 1
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Digital Image Processing 25
III.1.HistogramEqualization 2 conditions of T(r)
Single-valued (one-to-one relationship) guarantees that
the inverse transformation will exist
Monotonicity condition preserves the increasing order from
black to white in the output image thus it wont cause a
negative image
0 T(r) 1 for 0 r 1 guarantees that the output graylevels will be in the same range as the input levels.
The inverse transformation from s back to r is
r = Tr = T --11(s) ; 0(s) ; 0 ss 11
Digital Image Processing
26
III.1.HistogramEqualization
Let
pprr(r)(r) denote the PDF of random variable r
ppss(s(s)) denote the PDF of random variable s
Ifpprr(r)(r) and T(r)T(r) are known and TT--11(s)(s) satisfiescondition (a) thenppss(s(s)) can be obtained using a
formula :
ds
dr(r)p(s)p rs =
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Digital Image Processing 27
III.1.HistogramEqualization A transformation function is a cumulative distribution
function (CDF) of random variable r :
CDF is an integral of a probability function (always positive) is the
area under the function Thus, CDF is always single valued and monotonically increasing
Thus, CDF satisfies the condition (a)
We can use CDF as a transformation function
==r
r dw)w(p)r(Ts0
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28
III.1.HistogramEqualization
)r(p
dw)w(pdr
d
dr
)r(dT
dr
ds
r
r
r
=
=
=
0
10where1
1
=
=
=
s
)r(p)r(p
ds
dr)r(p)s(p
r
r
rs
Substitute and yield
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Digital Image Processing 29
III.1.HistogramEqualization Ps(s):
Asps(s) is a probability function, it must be zero outside
the interval [0,1] in this case because its integral over all
values of s must equal 1.
Calledps(s) as a uniform probability densitya uniform probability density functionfunction
ps(s) is always a uniform, independent of the form of
pr(r)
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III.1.HistogramEqualization
Discrete Transformation Function
The probability of occurrence of gray level in an image
is approximated by
The discrete version of transformation
=
=
==
==
k
j
j
k
j
jrkk
, ..., L-,where kn
n
)r(p)r(Ts
0
0
110
110 , ..., L-,where kn
n)r(p kkr ==
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Digital Image Processing 31
III.1.HistogramEqualization Thus, an output image is obtained by mapping
each pixel with level rrkk in the input image into a
corresponding pixel with level sskk in the output
image
In discrete space, it cannot be proved in general
that this discrete transformation will produce the
discrete equivalent of a uniform probability density
function, which would be a uniform histogram
Digital Image Processing
32
III.1.HistogramEqualization
Example before after Histogramequalization
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Digital Image Processing 33
III.1.HistogramEqualization Example
before after Histogramequalization
The quality is
not improvedmuch becausethe originalimage alreadyhas a broadengray-level scale
Digital Image Processing
34
III.Histogramprocessing
4x4 image
Gray scale = [0,9]histogram
0 1
1
2
2
3
3
4
4
5
5
6
6
7 8 9
No. of pixels
Gray level
2 3 3 2
4 2 4 3
3 2 3 5
2 4 2 4
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Digital Image Processing 35
III.1.HistogramEqualization Example
Gray Level 0 1 2 3 4 5 6 7 8 9
No.of pixels 0
16
16/
16
9s x 9
0 0 6 5 4 1 0 0 0
0 0 6 11 15 16 16 16 16
0 06 /
16
11 /
16
15 /
16
16 /
16
16/
16
16/
16
16/
16
0 03.3
3
6.1
6
8.4
89 9 9 9
=
k
j
jn0
=
=k
j
j
n
ns
0
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36
III.1.HistogramEqualization
Example
Output image
Gray scale = [0,9]
Histogram equalization
0 1
1
2
2
3
3
4
4
5
5
6
6
7 8 9
No. of pixels
Gray level
3 6 6 3
8 3 8 6
6 3 6 9
3 8 3 8
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Digital Image Processing 37
III.2.HistogramMatching Generate a processed image that has a specified
histogram
Let ( ) and ( ) denote the continous probability
density functions of the variables and . ( ) is the
specified probability density function.
Let be the random variable with the prob
r z
z
p r p z
r z p z
s
0
0
ability ( ) ( 1) ( )
Define a random variable with the probability
( ) ( 1) ( )
r
r
z
z
s T r L p w dw
z
G z L p t dt s
= =
= =
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38
III.2.HistogramMatching
0
0
( ) ( 1) ( )
( ) ( 1) ( )
r
r
z
z
s T r L p w dw
G z L p t dt s
= =
= =
[ ]1 1( ) ( )z G s G T r = =
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Digital Image Processing 39
III.2.HistogramMatching Procedure
1. Obtain pr(r) from the input image and then obtain the values of s
2. Use the specified PDF and obtain the transformation function
G(z)
3. Mapping from s to z
0( 1) ( )
r
rs L p w dw=
0( ) ( 1) ( )
z
zG z L p t dt s= =1( )z G s=
Digital Image Processing
40
III.2.HistogramMatching
Example
Assume an image has a gray level probability density functionpr(r) as shown.
0 1 2
1
2
Pr(r)
+=
elsewhere;0
1r;022 r)r(p r
10
=r
r dw)w(p
r
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Digital Image Processing 41
III.2.HistogramMatching Example
We would like to apply the histogram specification with thedesired probability density function pz(z) as shown.
0 1 2
1
2
Pz(z)
z
=elsewhere;0
1z;02z)z(pz
10
=z
z dw)w(p
Digital Image Processing
42
III.2.HistogramMatching
Example
0 1
1
s=T(r)
rrr
ww
dw)w(
dw)w(p)r(Ts
r
r
r
r
2
2
22
2
0
2
0
0
+=
+=
+=
==
1. Obtain the transformation function T(r)
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Digital Image Processing 45
III.2.HistogramMatching Procedure in discrete cases
1. Obtain pr(rj) from the input image and then obtain the values ofsk, round the value to the integer range [0, L-1]
2. Use the specified PDF and obtain the transformation functionG(zq), round the value to the integer range [0, L-1].
3. Mapping from sk to zq
0 0
( 1)( ) ( 1) ( )
k k
k k r j j
j j
Ls T r L p r n
MN= =
= = =
0
( ) ( 1) ( )q
q z i k
i
G z L p z s=
= =
1( )q kz G s=
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46
III.2.HistogramMatching
ExampleSuppose that a 3-bit image (L=8) of size 64 64 pixels (MN =4096) has the intensity distribution shown in the followingtable (on the left). Get the histogram transformation functionand make the output image with the specified histogram, listedin the table on the right.
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Digital Image Processing 47
III.2.HistogramMatching Example
Obtain the scaled histogram-equalized values,
Compute all the values of the transformation function G,
0 1 2 3 4
5 6 7
1, 3, 5, 6, 7,
7, 7, 7.
s s s s s
s s s
= = = = =
= = =
0
0
0
( ) 7 ( ) 0.00z jj
G z p z
=
= =
1 2
3 4
5 6
7
( ) 0.00 ( ) 0.00
( ) 1.05 ( ) 2.45
( ) 4.55 ( ) 5.95
( ) 7.00
G z G z
G z G z
G z G z
G z
= =
= =
= =
=
0
0 01 2
65
7
s0
s2 s3
s5 s6 s7
s1
s4
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48
III.2.HistogramMatching
Example 0 1 2 3 4
5 6 7
1, 3, 5, 6, 7,
7, 7, 7.
s s s s s
s s s
= = = = =
= = =
01
2
3
4
5
6
7
kr
0 3
1 4
2 5
3 6
4 7
5 7
6 7
7 7
k qr z
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Digital Image Processing 49
III.2.HistogramMatching Example
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50
III.2.HistogramMatching
Example
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Digital Image Processing 51
III.3.LocalHistogramProcessingDefine a neighborhood and move its center from pixel to pixel
At each location, the histogram of the points in the
neighborhood is computed. Either histogram equalization or
histogram specification transformation function is obtained
Map the intensity of the pixel centered in the neighborhood
Move to the next location and repeat the procedure
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52
III.3.LocalHistogramProcessing
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Digital Image Processing 53
III.4.UsingHistogramStatistics
1
0
( )L
i i
i
m r p r
=
= 1
0
( ) ( ) ( )L
n
n i i
i
u r r m p r
=
=
12 2
2
0
( ) ( ) ( )L
i i
i
u r r m p r
=
= =
1 1
0 0
1( , )
M N
x y
x yMN
= =
=
[ ]1 1
2
0 0
1( , )
M N
x y
f x y mMN
= =
=
Average Intensity
Variance
Digital Image Processing
54
III.4.UsingHistogramStatistics
1
0
Local average intensity
( )
denotes a neighborhood
xy xy
L
s i s i
i
xy
m r p r
s
=
=
12 2
0
Local variance
( ) ( )xy xy xy
L
s i s s i
i
r m p r
=
=
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Digital Image Processing 55
III.4.UsingHistogramStatistics
0 1 2
0 1 2
( , ), if and( , )
( , ), otherwise
: global mean; : global standard deviation
0.4; 0.02; 0.4; 4
xy xys G G s G
G G
E f x y m k m k kg x y
f x y
m
k k k E
=
= = = =
Digital Image Processing
56
IV.UsingArithmetic/LogicOperators
Arithmetic/Logic operations perform on pixel by
pixel basis between two or more images, except
NOT operation which perform only on a single
image
Logic operation performs on gray-level images,
the pixel values are processed as binary
numbers:
Light represents a binary 1, and
Dark represents a binary 0
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Digital Image Processing 57
IV.1.LogicOperators
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58
IV.2.ImageSubtraction
g(x,y) = f(x,y)g(x,y) = f(x,y)h(x,y)h(x,y)
Extraction of the differences between images
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Digital Image Processing 59
IV.3.ImageAveraging Consider a noisy image g(x,y) formed by the addition of noise (x,y)
to an original image f(x,y)
g(x,y) = f(x,y) + (x,y) If noise has zero mean and be uncorrelated then it can be shown that
if
),( yxg = image formed by averagingK different noisy images
=
=K
i
i yxgK
yxg1
),(1
),(
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60
IV.3.ImageAveraging
Then
),(2
),(2 1
yxyxg
K
=
= variances of g and ),(2
),(2 , yxyxg
if K increase, it indicates that the variability (noise) of the pixel ateach location (x,y) decreases.
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Digital Image Processing 63
V.SpatialFiltering
Example
M=3, N=3
3x3
Or M=5, N=5
5x5
Digital Image Processing
64
V.1.SmoothingFilters
Smoothing filters are used for blurring and for noise reduction
Blurring is used in removal of small details and bridging of small
gaps in lines or curves
Smoothing spatial filters include linear filters and nonlinear filters.
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Digital Image Processing 65
V.1.SmoothingFilters Linear filters
The general implementation for filtering an M N image
with a weighted averaging filter of size m n is given
( , ) ( , )
( , )( , )
where 2 1
a b
s a t b
a b
s a t b
w s t f x s y t
g x yw s t
m a
= =
= =
+ +
=
= +
, 2 1.n b= +
Digital Image Processing
66
V.1.SmoothingFilters
Linear filters
Averaging Filter Masks
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Digital Image Processing 67
V.1.SmoothingFilters Example
Averaging Filter Masks
a). original image 500x500 pixel
b). - f). results of smoothing with
square averaging filter masks of
size n = 3, 5, 9, 15 and 35,
respectively.
Note:
Big mask is used to eliminate small objects
from an image.
The size of the mask establishes the
relative size of the objects that will be
blended with the background.
a b
c d
e f
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68
V.1.SmoothingFilters
original image result after smoothing with15x15 averaging mask
result of thresholding
we can see that the result after smoothing and thresholding, theremains are the largest and brightest objects in the image.
Example
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Digital Image Processing 69
V.1.SmoothingFilters Order-Statistics Filters (Nonlinear Filters)
The response is based on ordering (ranking) the pixels
contained in the image area encompassed by the filter
Example
Median filter : R = median{zk |k = 1,2,,n x n}
Max filter : R = max{zk |k = 1,2,,n x n}
Min filter : R = min{zk |k = 1,2,,n x n}
Note: n x n is the size of the mask
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70
V.1.SmoothingFilters
Median Filter Median: X= {x1,x2,, x2b+1}, x is called the median of X if x
greater than or equal b elements and less than or equal b other
elements in X;
For example: X={3,2,3,2,3,4,5,6,5} (b=4) -> Median is 3
Peplaces the value of a pixel by the median of the gray levels in
the neighborhood of that pixel (the original value of the pixel is
included in the computation of the median)
Quite popular because for certain types of random noise (impulseimpulse
noisenoise salt and pepper noisesalt and pepper noise) , they provide excellent noiseprovide excellent noise--
reduction capabilitiesreduction capabilities, with considering less blurring than linearless blurring than linear
smoothing filters of similar sizesmoothing filters of similar size
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Digital Image Processing 71
V.1.SmoothingFilters Median Filter
Example
Digital Image Processing
72
V.2.SharpeningSpatialFilters
Foundation
Laplacian Operator
Unsharp Masking and Highboost Filtering
Using First-Order Derivatives for Nonlinear Image
Sharpening - The Gradient
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Digital Image Processing 75
Laplace Operator
V.3.MethodbasedonLaplaceOperator
The second-order isotropic derivative operator is the Laplacian
for a function (image) f(x,y)2 2
2
2 2
f ff
x y
= +
2
2( 1, ) ( 1, ) 2 ( , )
ff x y f x y f x y
x
= + +
22
( , 1) ( , 1) 2 ( , )f
f x y f x y f x yy
= + +
2 ( 1, ) ( 1, ) ( , 1) ( , 1)
- 4 ( , )
f f x y f x y f x y f x y
f x y
= + + + + +
Digital Image Processing
76
Masks
V.3.MethodbasedonLaplaceOperator
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Digital Image Processing 77
Image sharpening in the way of using the
Laplacian:
where g(x,y) sharpened image
V.3.MethodbasedonLaplaceOperator
Digital Image Processing
78
Example
V.3.MethodbasedonLaplaceOperator
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Digital Image Processing 79
V.4. Unsharp Masking and Highboost
Filtering
Digital Image Processing
80
V.4. Unsharp Masking and Highboost
Filtering
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Digital Image Processing 81
V.4. Unsharp Masking and Highboost
Filtering
Digital Image Processing
82
V.4. Unsharp Masking and Highboost
Filtering Example
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Digital Image Processing 83
V.4. Unsharp Masking and Highboost
Filtering Example
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84
V.5. Method based on First-Order
Derivatives(Gradient)
For function ( , ), the gradient of at coordinates ( , )
is defined as
grad( ) x
y
f x y f x y
f
g xf f
fg
y
=
2 2
The of vector , denoted as ( , )
( , ) mag( ) x y
magnitude f M x y
M x y f g g
= = +
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Digital Image Processing 85
V.5. Method based on First-Order
Derivatives(Gradient)
2 2
The of vector , denoted as ( , )
( , ) mag( ) x y
magnitude f M x y
M x y f g g
= = +
( , ) | | | |x y
x y g g +
z1 z2 z3
z4 z5 z6
z7 z8 z9
8 5 6 5( , ) | | | |x y z z z z= +
Digital Image Processing
86
V.5. Method based on First-Order
Derivatives(Gradient)
z1 z2 z3
z4 z5 z6
z7 z8 z9
9 5 8 6
Roberts Cross-gradient Operators
( , ) | | | |M x y z z z z +
7 8 9 1 2 3
3 6 9 1 4 7
Sobel Operators
( , ) | ( 2 ) ( 2 ) |
| ( 2 ) ( 2 ) |
M x y z z z z z z
z z z z z z
+ + + +
+ + + + +
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Digital Image Processing 87
V.5. Method based on First-Order
Derivatives(Gradient)
Digital Image Processing
88
V.5. Method based on First-Order
Derivatives(Gradient)
Example
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Intensity transformation and spatial filtering
45
Digital Image Processing 89
V.6. Combining Spatial Enhancement
Methods
Solve:
1. Laplacian to highlight fine detail
2. Gradient to enhance prominent edges
3. Gray-level transformation to increase the
dynamic range of gray levels
Digital Image Processing
90
V.6. Combining Spatial Enhancement
Methods
Exampe
8/12/2019 Lecture 03,04,05 - Intensity transformation and Spatial Filtering.pdf
46/46
Intensity transformation and spatial filtering
Digital Image Processing 91
V.6. Combining Spatial Enhancement
Methods
Exampe
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