Codeforces Round 277 (Div. 2) |
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Finished |

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B. OR in Matrix

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputLet's define logical *OR* as an operation on two logical values (i. e. values that belong to the set {0, 1}) that is equal to 1 if either or both of the logical values is set to 1, otherwise it is 0. We can define logical *OR* of three or more logical values in the same manner:

where is equal to 1 if some *a*_{i} = 1, otherwise it is equal to 0.

Nam has a matrix *A* consisting of *m* rows and *n* columns. The rows are numbered from 1 to *m*, columns are numbered from 1 to *n*. Element at row *i* (1 ≤ *i* ≤ *m*) and column *j* (1 ≤ *j* ≤ *n*) is denoted as *A*_{ij}. All elements of *A* are either 0 or 1. From matrix *A*, Nam creates another matrix *B* of the same size using formula:

.

(*B*_{ij} is *OR* of all elements in row *i* and column *j* of matrix *A*)

Nam gives you matrix *B* and challenges you to guess matrix *A*. Although Nam is smart, he could probably make a mistake while calculating matrix *B*, since size of *A* can be large.

Input

The first line contains two integer *m* and *n* (1 ≤ *m*, *n* ≤ 100), number of rows and number of columns of matrices respectively.

The next *m* lines each contain *n* integers separated by spaces describing rows of matrix *B* (each element of *B* is either 0 or 1).

Output

In the first line, print "NO" if Nam has made a mistake when calculating *B*, otherwise print "YES". If the first line is "YES", then also print *m* rows consisting of *n* integers representing matrix *A* that can produce given matrix *B*. If there are several solutions print any one.

Examples

Input

2 2

1 0

0 0

Output

NO

Input

2 3

1 1 1

1 1 1

Output

YES

1 1 1

1 1 1

Input

2 3

0 1 0

1 1 1

Output

YES

0 0 0

0 1 0

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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