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 r[110] = {0}, c[110] = {0}, b[110][110];
int main() {
  int n, m, x;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      scanf("%d", b[i] + j);
      if (b[i][j]) ++r[i], ++c[j];
    }
  int ok = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j) {
      if (r[i] == m && c[j] == n) {
        for (int k = 1; k <= n; ++k)
          if (!b[k][j])
            ok = 0;
          else
            b[k][j] = -1;
        for (int k = 1; k <= m; ++k)
          if (!b[i][k])
            ok = 0;
          else
            b[i][k] = -1;
      }
    }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= m; ++j)
      if (b[i][j] && b[i][j] > 0) ok = 0;
  if (ok) {
    puts("YES");
    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j)
        if (r[i] == m && c[j] == n)
          printf("1%c", j == m ? '\n' : ' ');
        else
          printf("0%c", j == m ? '\n' : ' ');
  } else
    puts("NO");
  return 0;
}