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;
void AWAD() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
}
bool cmp(long long int a, long long int b) {
  if (a > b) return 1;
  return false;
}
int mata[102][102], matb[102][102], matc[102][102];
int main() {
  AWAD();
  for (int i = 0; i < 102; i++) {
    for (int j = 0; j < 102; j++) {
      mata[i][j] = 1;
    }
  }
  long long int row, col;
  cin >> row >> col;
  for (int i = 0; i < row; i++) {
    for (int j = 0; j < col; j++) {
      cin >> matb[i][j];
      if (matb[i][j] == 0) {
        for (int k = 0; k < col; k++) mata[i][k] = 0;
        for (int k = 0; k < row; k++) mata[k][j] = 0;
      }
    }
  }
  bool chick = true;
  for (int i = 0; i < row; i++) {
    for (int j = 0; j < col; j++) {
      int wee = 0;
      if (mata[i][j] == 1) {
        for (int k = 0; k < col; k++) matc[i][k] = 1;
        for (int k = 0; k < row; k++) matc[k][j] = 1;
      }
    }
  }
  for (int i = 0; i < row; i++)
    for (int j = 0; j < col; j++)
      if (matc[i][j] != matb[i][j]) {
        chick = false;
        break;
      }
  if (!chick) {
    cout << "NO\n";
  } else {
    cout << "YES" << endl;
    for (int i = 0; i < row; i++) {
      for (int j = 0; j < col; j++) {
        cout << mata[i][j];
        if (j + 1 != col) cout << " ";
      }
      cout << endl;
    }
  }
  return 0;
}