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>
const long long N = 110;
const long long mod = 1e9 + 7;
using namespace std;
int a[N][N];
int raw[N], arrange[N];
int main() {
  int n, m, i, j;
  scanf("%d%d", &n, &m);
  for (i = 1; i <= n; i++) raw[i] = 1;
  for (i = 1; i <= m; i++) arrange[i] = 1;
  for (i = 1; i <= n; i++)
    for (j = 1; j <= m; j++) {
      scanf("%d", &a[i][j]);
      if (a[i][j] == 0) {
        raw[i] = arrange[j] = 0;
      }
    }
  int t = 0;
  for (i = 1; i <= n; i++) {
    if (raw[i] == 0) t++;
  }
  if (t == n) {
    for (i = 1; i <= m; i++) arrange[i] = 0;
  }
  t = 0;
  for (i = 1; i <= m; i++) {
    if (arrange[i] == 0) t++;
  }
  if (t == m) {
    for (i = 1; i <= n; i++) raw[i] = 0;
  }
  int flag = 1;
  for (i = 1; i <= n && flag; i++)
    for (j = 1; j <= m && flag; j++) {
      if (a[i][j]) {
        if (!raw[i] && !arrange[j]) {
          flag = 0;
          break;
        }
      }
    }
  if (flag) {
    printf("YES\n");
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        printf("%d", (raw[i] && arrange[j]) != 0 ? 1 : 0);
        if (j == m)
          printf("\n");
        else
          printf(" ");
      }
    }
  } else
    printf("NO\n");
  return 0;
}