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;
long long int power(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return power(x * x, n / 2);
  else
    return x * power(x * x, (n - 1) / 2);
}
long long int highestPowerof2(long long int n) {
  long long int p = (long long int)log2(n);
  return (long long int)pow(2, p);
}
long long int powermod(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k, count;
  bool ok = false;
  cin >> n >> m;
  long long int a[n][m];
  long long int arr[n][m];
  memset(arr, -1, sizeof(arr));
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) cin >> a[i][j];
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (a[i][j] == 0) {
        for (k = 0; k < n; k++) arr[k][j] = 0;
        for (k = 0; k < m; k++) arr[i][k] = 0;
      }
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (arr[i][j] == -1) arr[i][j] = 1;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      count = 0;
      if (a[i][j] == 1) {
        for (k = 0; k < n; k++)
          if (arr[k][j] == 1) count++;
        for (k = 0; k < m; k++)
          if (arr[i][k] == 1) count++;
        if (count == 0) {
          ok = true;
          break;
        }
      } else if (a[i][j] == 0) {
        for (k = 0; k < n; k++)
          if (arr[k][j] == 1) count++;
        for (k = 0; k < m; k++)
          if (arr[i][k] == 1) count++;
        if (count != 0) {
          ok = true;
          break;
        }
      }
    }
    if (ok) break;
  }
  if (ok)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (i = 0; i < n; i++) {
      for (j = 0; j < m; j++) {
        cout << arr[i][j] << " ";
      }
      cout << endl;
    }
  }
}