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 m, n, i, j, a[110][110], b[110][110], k;
int main() {
  scanf("%d%d", &m, &n);
  for (i = 1; i <= m; i++)
    for (j = 1; j <= n; j++) {
      a[i][j] = 1;
      scanf("%d", &b[i][j]);
    }
  for (i = 1; i <= m; i++)
    for (j = 1; j <= n; j++) {
      if (b[i][j] == 0) {
        for (k = 1; k <= n; k++) a[i][k] = 0;
        for (k = 1; k <= m; k++) a[k][j] = 0;
      }
    }
  for (i = 1; i <= m; i++)
    for (j = 1; j <= n; j++) {
      int tmp = 0;
      for (k = 1; k <= n; k++) tmp |= a[i][k];
      for (k = 1; k <= m; k++) tmp |= a[k][j];
      if (tmp != b[i][j]) {
        printf("NO\n");
        return 0;
      }
    }
  printf("YES\n");
  for (i = 1; i <= m; i++) {
    for (j = 1; j < n; j++) printf("%d ", a[i][j]);
    printf("%d\n", a[i][n]);
  }
}