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;
template <class T>
T bigmod(T b, T p, T m) {
  if (!p) return 1;
  T res = (bigmod(b, p / 2, m)) % m;
  res = (res * res) % m;
  if (p & 1) res = (res * b) % m;
  return res % m;
}
template <class T>
T modinv(T b, T m) {
  return bigmod(b, m - 2, m);
}
const long long INF = 1000000007;
const long long MOD = 1000000007;
const int MAX = 1000005;
const int sz = 100005;
int main() {
  int a[105][105], b[105][105], m, n, i, j, k, l, p, f, F;
  while (cin >> m >> n) {
    memset(a, -1, sizeof(a));
    for (i = 1; i <= m; i++)
      for (j = 1; j <= n; j++) cin >> b[i][j];
    for (i = 1; i <= m; i++) {
      for (j = 1; j <= n; j++) {
        if (b[i][j] == 0) {
          for (k = 1; k <= n; k++) a[i][k] = 0;
          for (k = 1; k <= m; k++) a[k][j] = 0;
        }
      }
    }
    for (i = 1; i <= m; i++)
      for (j = 1; j <= n; j++) a[i][j] *= -1;
    F = 0;
    for (i = 1; i <= m; i++) {
      for (j = 1; j <= n; j++) {
        if (b[i][j]) {
          f = 1;
          for (k = 1; k <= n; k++)
            if (a[i][k]) f = 0;
          if (f)
            for (k = 1; k <= m; k++)
              if (a[k][j]) f = 0;
          if (f) {
            F = 1;
            break;
          }
        }
      }
      if (F) break;
    }
    if (F)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (i = 1; i <= m; i++) {
        for (j = 1; j <= n; j++) cout << a[i][j] << " ";
        cout << endl;
      }
    }
  }
  return 0;
}