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>
#pragma GCC optimize("Ofast,inline,unroll-loops,fast-math")
#pragma GCC optimize("O3")
using namespace std;
using namespace std::chrono;
long long mod_exp(long long x, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = (res * x) % 1000000007;
    x = (x * x) % 1000000007;
    n >>= 1;
  }
  return res;
}
void inout() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  vector<vector<int>> b(n + 5, vector<int>(m + 5)),
      a(n + 5, vector<int>(m + 5, 1));
  for (int i = 0; i < (int)n; ++i)
    for (int j = 0; j < (int)m; ++j) cin >> b[i][j];
  for (int i = 0; i < (int)n; ++i) {
    for (int j = 0; j < (int)m; ++j) {
      if (!b[i][j]) {
        for (int k = 0; k < (int)m; ++k) {
          a[i][k] = 0;
        }
        for (int p = 0; p < (int)n; ++p) {
          a[p][j] = 0;
        }
      }
    }
  }
  bool flag = 0;
  for (int i = 0; i < (int)n; ++i) {
    for (int j = 0; j < (int)m; ++j) {
      bool x = 0;
      for (int k = 0; k < (int)m; ++k) {
        x |= a[i][k];
      }
      for (int p = 0; p < (int)n; ++p) {
        x |= a[p][j];
      }
      if (b[i][j] != x) {
        flag = 1;
        break;
      }
    }
  }
  if (flag)
    cout << "NO";
  else {
    cout << "YES" << '\n';
    for (int i = 0; i < (int)n; ++i) {
      for (int j = 0; j < (int)m; ++j) cout << a[i][j] << " ";
      cout << '\n';
    }
  }
  return 0;
}