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 s[101][101], f[101][101];
int main() {
  int n, m;
  cin >> n >> m;
  for (long long i = (0); i < (n); i++) {
    for (long long j = (0); j < (m); j++) {
      cin >> s[i][j];
    }
  }
  int sum1 = 0, sum2 = 0, sum3 = 0;
  for (long long i = (0); i < (n); i++) {
    for (long long j = (0); j < (m); j++) {
      if (s[i][j] == 1) {
        if (n + m < 3) {
          sum3 = 1;
        } else {
          sum3 = 0;
        }
        if (i != 0) {
          sum3 += s[i - 1][j];
        }
        if (j != 0) {
          sum3 += s[i][j - 1];
        }
        if (i < n) {
          sum3 += s[i + 1][j];
        }
        if (j < m) {
          sum3 += s[i][j + 1];
        }
        if (!sum3) {
          cout << "NO";
          return 0;
        }
        sum2++;
        bool bo = true;
        for (long long k = (0); k < (n); k++) {
          if (!s[k][j]) {
            bo = false;
          }
        }
        for (long long k = (0); k < (m); k++) {
          if (!s[i][k]) {
            bo = false;
          }
        }
        if (bo) {
          f[i][j] = 1;
          sum1 += n + m - 1;
        } else {
          f[i][j] = 0;
        }
      }
    }
  }
  if (sum1 < sum2) {
    cout << "NO";
  } else {
    cout << "YES" << endl;
    for (long long i = (0); i < (n); i++) {
      for (long long j = (0); j < (m); j++) {
        cout << f[i][j] << " ";
      }
      cout << endl;
    }
  }
}