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 cmpfunc(const void *a, const void *b) { return (*(int *)a - *(int *)b); }
long long int gcd(long long int m, long long int n) {
  if (n == 0) return m;
  return gcd(n, m % n);
}
long long int fastpow(long long int a, long long int b, long long int m) {
  long long int r = 1;
  while (b > 0) {
    if (b % 2 == 1) r = (r * a) % m;
    b = b >> 1;
    a = (a * a) % m;
  }
  return r;
}
int prime(long long int x) {
  if (x == 1) return 0;
  if (x <= 3) return 1;
  if (x % 6 == 1 || x % 6 == 5) {
    long long int y = sqrt(x), i;
    for (i = 2; i <= y; i++)
      if (x % i == 0) return 0;
    return 1;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int m, n, i, j, ans1 = 1, ans2 = 1, flag = 0, bc = 0, k, haha = 0;
  int A[100][100] = {0}, B[100][100];
  cin >> m >> n;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      cin >> B[i][j];
    }
  }
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      if (B[i][j] == 1) {
        bc = 1;
        for (k = 0; k < n; k++) {
          ans1 = ans1 & B[i][k];
        }
        for (k = 0; k < m; k++) {
          ans2 = ans2 & B[k][j];
        }
        if (ans1 == 0 && ans2 == 0) {
          flag = 1;
          break;
        }
        if (ans1 == 1 && ans2 == 1 && flag == 0) {
          A[i][j] = 1;
          haha = 1;
        }
        ans1 = 1;
        ans2 = 1;
      }
    }
    if (flag == 1) break;
  }
  if (flag == 1) {
    cout << "NO\n";
  } else if (bc == 1 && haha == 0) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        cout << A[i][j] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}