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;
const long long maxi = 1e17;
const int MAXM = 1e2 + 5;
const int MAXN = 5e2 + 5;
int dp[MAXM][MAXM], ro[MAXM][MAXM], co[MAXM][MAXM], a[MAXM][MAXM],
    ans[MAXM][MAXM];
int ff = 0, fff = 0;
bool check(int n, int m) {
  int i, j, k;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      if (dp[i][j] == 1) {
        ff = 0;
        for (k = 1; k <= n; k++) {
          if (ans[k][j] == 1) ff = 1;
        }
        for (k = 1; k <= m; k++) {
          if (ans[i][k] == 1) ff = 1;
        }
        if (ff != 1) {
          fff = 1;
        }
      } else {
        ff = 0;
        for (k = 1; k <= n; k++) {
          if (ans[k][j] == 1) ff = 1;
        }
        for (k = 1; k <= m; k++) {
          if (ans[i][k] == 1) ff = 1;
        }
        if (ff == 1) {
          fff = 1;
        }
      }
    }
  }
  if (fff == 1)
    return false;
  else
    return true;
}
int main() {
  int n, m, i, j, k;
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      cin >> dp[i][j];
      ro[i][j] = ro[i][j - 1] + dp[i][j];
    }
  }
  for (i = 1; i <= m; i++) {
    for (j = 1; j <= n; j++) {
      co[j][i] = co[j - 1][i] + dp[j][i];
    }
  }
  int f = 0;
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      if (co[n][j] != n && ro[i][m] != m && dp[i][j] == 1) {
        f = 1;
      }
    }
  }
  if (f) {
    cout << "NO" << endl;
    return 0;
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= m; j++) {
      if (co[n][j] == n && ro[i][m] == m) {
        ans[i][j] = 1;
      } else {
        ans[i][j] = 0;
      }
    }
  }
  if (check(n, m)) {
    cout << "YES" << endl;
    for (i = 1; i <= n; i++) {
      for (j = 1; j <= m; j++) {
        cout << ans[i][j] << " ";
      }
      cout << endl;
    }
  } else {
    cout << "NO" << endl;
  }
  return 0;
}