Problem: 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 logical OR of three or more logical values in the same manner:

<image> where <image> 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:

<image>.

(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.

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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int r[110], l[110], val[110][110], num[110][110];
int main() {
  int T, t, n, m, i, j, k;
  bool flag;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) r[i] = 1;
  for (j = 1; j <= m; j++) l[j] = 1;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) {
      scanf("%d", &num[i][j]);
      if (num[i][j] == 0) r[i] = l[j] = 0;
    }
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (r[i] == 1 && l[j] == 1)
        val[i][j] = 1;
      else
        val[i][j] = 0;
  for (i = 1; i <= n; i++) {
    r[i] = 0;
    for (j = 1; j <= m; j++) r[i] += val[i][j];
  }
  for (i = 1; i <= m; i++) {
    l[i] = 0;
    for (j = 1; j <= n; j++) l[i] += val[j][i];
  }
  flag = true;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++)
      if (num[i][j] == 1 && r[i] + l[j] == 0) flag = false;
  if (!flag)
    printf("NO\n");
  else {
    printf("YES\n");
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) printf("%d ", val[i][j]);
      printf("\n");
    }
  }
}