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>
int main() {
  int n, m;
  int ones = 0;
  scanf("%d %d", &n, &m);
  int input[n][m];
  int i = 0;
  while (i < n) {
    int j = 0;
    while (j < m) {
      scanf("%d", &input[i][j]);
      if (input[i][j] == 1) {
        ones++;
      }
      j++;
    }
    i++;
  }
  static int rows[200];
  static int columns[200];
  int row = 0;
  int column = 0;
  int k = 0;
  while (k < n) {
    int rcheck = 0;
    int x = 0;
    while (x < m) {
      if (input[k][x] != 1) {
        rcheck = 1;
        break;
      }
      x++;
    }
    if (rcheck == 0) {
      rows[k] = 1;
      row++;
    }
    k++;
  }
  int y = 0;
  while (y < m) {
    int ccheck = 0;
    int x = 0;
    while (x < n) {
      if (input[x][y] != 1) {
        ccheck = 1;
        break;
      }
      x++;
    }
    if (ccheck == 0) {
      columns[y] = 1;
      column++;
    }
    y++;
  }
  int ans = m * row + column * (n - row);
  if ((ans == ones && row > 0 && column > 0) ||
      (ans == ones && row == 0 && column == 0)) {
    printf("YES\n");
    int abc = 0;
    while (abc < n) {
      int def = 0;
      while (def < m) {
        if (rows[abc] == 1 && columns[def] == 1) {
          printf("1 ");
        } else {
          printf("0 ");
        }
        def++;
      }
      printf("\n");
      abc++;
    }
  } else {
    printf("NO");
  }
  return 0;
}