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;
bool sortbysec(const pair<long long int, long long int> &a,
               const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n, m, i, j, z = 0;
  cin >> n >> m;
  vector<vector<long long int>> v(n, vector<long long int>(m, 0)),
      ans(n, vector<long long int>(m, 1));
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      cin >> v[i][j];
      if (v[i][j] == 0) z++;
    }
  }
  if (z == n * m) {
    cout << "YES" << endl;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) cout << v[i][j] << " ";
      cout << endl;
    }
    return 0;
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (v[i][j] == 0) {
        long long int k;
        for (k = 0; k < n; k++) ans[k][j] = 0;
        for (k = 0; k < m; k++) ans[i][k] = 0;
      }
    }
  }
  long long int f = 0;
  for (i = 0; i < n; i++) {
    f = 1;
    for (j = 0; j < m; j++) {
      if (v[i][j] == 1) {
        f = 0;
        long long int k;
        for (k = 0; k < n; k++) {
          if (ans[k][j] == 1) {
            f = 1;
            break;
          }
        }
        if (f == 0) {
          for (k = 0; k < m; k++) {
            if (ans[i][k] == 1) {
              f = 1;
              break;
            }
          }
        }
      }
      if (f == 0) break;
    }
    if (f == 0) break;
  }
  if (f == 0)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) cout << ans[i][j] << " ";
      cout << endl;
    }
  }
  return 0;
}