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;
inline long long int Power(int b, int p) {
  long long int ret = 1;
  for (int i = 1; i <= p; i++) ret *= b;
  return ret;
}
const int MOD = 1e9 + 7;
const int SIZE = 4e6 + 10;
const int MAX = 1e9 + 1;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  int flag1 = 0, flag2 = 0;
  int a[n][m], b[n][m];
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      a[i][j] = 1;
    }
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      int(x);
      scanf("%d", &x);
      b[i][j] = x;
      if (x == 0) {
        for (int k = (0); k < (n); ++k) a[k][j] = 0;
        for (int l = (0); l < (m); ++l) a[i][l] = 0;
      }
    }
  }
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      int tmp = a[i][j];
      for (int k = (0); k < (n); ++k) {
        tmp = tmp | a[k][j];
      }
      for (int l = (0); l < (m); ++l) {
        tmp = tmp | a[i][l];
      }
      if (tmp != b[i][j]) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  cout << "YES\n";
  for (int i = (0); i < (n); ++i) {
    for (int j = (0); j < (m); ++j) {
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}