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")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
long long MOD = 998244353;
double eps = 1e-12;
long long arr1[100001];
long long arr2[100001];
long long arr2D[1001][1001];
vector<long long> vec;
void solve() {
  long long n, m, a, b, c, d;
  string s;
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) cin >> arr2D[i][j];
  }
  int ans2D[1001][1001];
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      if (arr2D[i][j] == 0) {
        ans2D[i][j] = 0;
        continue;
      } else {
        int flag = 0;
        for (long long p = 0; p < n; p++) {
          if (arr2D[p][j] == 0) {
            flag += 1;
            break;
          }
        }
        for (long long p = 0; p < m; p++) {
          if (arr2D[i][p] == 0) {
            flag += 1;
            break;
          }
        }
        if (flag == 2) {
          cout << "NO" << endl;
          return;
        }
        if (flag == 1) {
          ans2D[i][j] = 0;
          continue;
        }
        ans2D[i][j] = 1;
        continue;
      }
    }
  }
  int reqarr[1001][1001];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      int myor = 0;
      for (long long p = 0; p < n; p++) myor |= ans2D[p][j];
      for (long long p = 0; p < m; p++) myor |= ans2D[i][p];
      reqarr[i][j] = myor;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) {
      if (arr2D[i][j] != reqarr[i][j]) {
        cout << "NO" << endl;
        return;
      }
    }
  }
  cout << "YES" << endl;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < m; j++) cout << ans2D[i][j] << " ";
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  return 0;
}