cf 486B

Let'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 logicalOR of three or more logical values in the same manner:

$cf 486B$ where $cf 486B$ is equal to 1 if some ai = 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 Aij. All elements of A are either 0 or 1. From matrix A, Nam creates another matrix B of the same size using formula:

$cf 486B$ .

(Bij 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.

Sample test(s)

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

可以去死了，，，

A矩阵初始化全部赋值为1，然后 B矩阵为0的位置 直接 处理一遍A矩阵。注意当B矩阵元素为一时，需要判断一下A矩阵是否满足啊啊啊

#include<iostream>
#include<cstdio>
using namespace std;
int a[110][110],b[110][110],n,m;
int main()
{
      scanf("%d%d",&n,&m);
      for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            {
                  scanf("%d",&a[i][j]);
                  b[i][j]=1;
            }
      for(int i=1;i<=n;i++)
      {
            for(int j=1;j<=m;j++)
            {
                  if(a[i][j]==0)
                  {
                        for(int k=1;k<=n;k++)
                              b[k][j]=0;
                        for(int k=1;k<=m;k++)
                              b[i][k]=0;
                  }
            }
      }
      for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
            {
                  if(a[i][j]==1)
                  {
                        bool flag=0;
                        for(int k=1;k<=n;k++)
                              flag|=b[k][j];
                        for(int k=1;k<=m;k++)
                              flag|=b[i][k];
                        if(!flag)
                        {
                              printf("NO
");
                              return 0;
                        }
                  }
            }
      printf("YES
");
      for(int i=1;i<=n;i++)
      {
            for(int j=1;j<m;j++)
                  printf("%d ",b[i][j]);
            printf("%d
",b[i][m]);
      }
      return 0;
}

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