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;
void solve() {
  long long int n, m, flag = 0;
  cin >> n >> m;
  long long int a[n][m];
  long long int b[n][m];
  long long int c[n][m];
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      cin >> b[i][j];
      a[i][j] = 1;
      c[i][j] = 0;
    }
  }
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      if (b[i][j] == 0) {
        long long int k = 0;
        while (k < m) {
          a[i][k] = 0;
          k++;
        }
        k = 0;
        while (k < n) {
          a[k][j] = 0;
          k++;
        }
      }
    }
  }
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      long long int k = 0;
      while (k < m) {
        c[i][j] = c[i][j] | a[i][k];
        k++;
      }
      k = 0;
      while (k < n) {
        c[i][j] = c[i][j] | a[k][j];
        k++;
      }
    }
  }
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < m; j++) {
      if (b[i][j] != c[i][j]) {
        flag = 1;
        break;
      }
      if (flag == 1) break;
    }
  }
  if (flag == 0) {
    cout << "YES"
         << "\n";
    for (long long int i = 0; i < n; i++) {
      for (long long int j = 0; j < m; j++) cout << a[i][j] << " ";
      cout << "\n";
    }
  } else
    cout << "NO"
         << "\n";
}
int32_t main() { solve(); }