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;
const int siz = 1e5 + 7;
const int siz2 = 2e5 + 7;
int n, m;
int a[111][111], b[111][111];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> m >> n;
  for (int i = 0; i < m; ++i)
    for (int j = 0; j < n; ++j) cin >> a[i][j];
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      int Cnt = 0;
      for (int k = 0; k < n; ++k) Cnt += a[i][k];
      for (int k = 0; k < m; ++k) Cnt += a[k][j];
      if (Cnt == n + m) b[i][j] = 1;
    }
  }
  bool Flag = false;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) {
      int Cnt = 0;
      for (int k = 0; k < n; ++k) Cnt |= b[i][k];
      for (int k = 0; k < m; ++k) Cnt |= b[k][j];
      if (a[i][j] != Cnt) Flag = true;
    }
  }
  if (Flag) return cout << "NO", 0;
  cout << "YES" << endl;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < n; ++j) cout << b[i][j] << " ";
    cout << endl;
  }
  return 0;
}