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 num[100 + 5][100 + 5], ans[100 + 5][100 + 5];
int row[100 + 5], col[100 + 5];
int main() {
  int n, m;
  while (scanf("%d%d", &m, &n) != EOF) {
    int i, j, k;
    int ii, jj, kk;
    memset(row, 0, sizeof(row));
    memset(col, 0, sizeof(col));
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        scanf("%d", &num[i][j]);
        row[i] += num[i][j];
        col[j] += num[i][j];
      }
    }
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        ans[i][j] = 1;
      }
    }
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        if (num[i][j] == 0) {
          for (ii = 0; ii < m; ii++) ans[ii][j] = 0;
          for (jj = 0; jj < n; jj++) ans[i][jj] = 0;
        }
      }
    }
    bool ok = true;
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        int tp = 0;
        for (ii = 0; ii < m; ii++) tp |= ans[ii][j];
        for (jj = 0; jj < n; jj++) tp |= ans[i][jj];
        if (tp != num[i][j]) {
          ok = false;
          break;
        }
      }
      if (!ok) break;
    }
    if (ok) {
      printf("YES\n");
      for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++) {
          if (j) printf(" ");
          printf("%d", ans[i][j]);
        }
        printf("\n");
      }
    } else
      printf("NO\n");
  }
  return 0;
}