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;
int main() {
  int n, m;
  cin >> n >> m;
  bool A[n][m], B[n][m];
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      cin >> B[i][j];
    }
  }
  memset(A, 1, sizeof(A));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (!B[i][j]) {
        for (int k = 0; k < m; ++k) A[i][k] = 0;
        for (int k = 0; k < n; ++k) A[k][j] = 0;
      }
    }
  }
  bool C[n][m], row[n], col[m];
  memset(row, 0, sizeof(row));
  memset(col, 0, sizeof(col));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      if (!j)
        row[i] = A[i][j];
      else
        row[i] = row[i] || A[i][j] || A[i][j - 1];
    }
  }
  for (int j = 0; j < m; ++j) {
    for (int i = 0; i < n; ++i) {
      if (!i)
        col[j] = A[i][j];
      else
        col[j] = col[j] || A[i][j] || A[i - 1][j];
    }
  }
  bool ok = 1;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      C[i][j] = (row[i] or col[j]);
      if (C[i][j] != B[i][j]) ok = 0;
    }
  }
  if (!ok)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < m; ++j) {
        cout << A[i][j] << " ";
      }
      cout << endl;
    }
  }
}