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>
int is_row(int *arr, int r, int m, int n) {
  int *t = arr + n * r;
  for (int i = 0; i < n; i++)
    if (*(t + i) == 0) return 0;
  return 1;
}
int is_col(int *arr, int c, int m, int n) {
  for (int i = 0; i < m; i++)
    if (*(arr + i * n + c) == 0) return 0;
  return 1;
}
void assign_row(int *arr, int r, int m, int n, int x) {
  int *t = arr + n * r;
  for (int i = 0; i < n; i++) *(t + i) = x;
}
void assign_col(int *arr, int c, int m, int n, int x) {
  for (int i = 0; i < m; i++) *(arr + i * n + c) = x;
}
int main() {
  int m, n, i, j;
  scanf("%d %d", &m, &n);
  int arr[m][n], temp[m][n], t1[m][n];
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      scanf("%d", &arr[i][j]);
      temp[i][j] = 1;
      t1[i][j] = 0;
    }
  }
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      if (arr[i][j] == 0) {
        assign_col((int *)temp, j, m, n, 0);
        assign_row((int *)temp, i, m, n, 0);
      }
    }
  }
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      int flag = 0;
      for (int k = 0; k < n; k++) {
        flag = flag || temp[i][k];
        if (flag == 1) {
          t1[i][j] = 1;
          if (arr[i][j] != t1[i][j]) {
            printf("NO");
            return 0;
          }
          break;
        }
      }
      if (t1[i][j] == 1) continue;
      for (int k = 0; k < m; k++) {
        flag = 0;
        flag = flag || temp[k][j];
        if (flag == 1) {
          t1[i][j] = 1;
          if (arr[i][j] != t1[i][j]) {
            printf("NO");
            return 0;
          }
          break;
        }
      }
    }
  }
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++)
      if (arr[i][j] != t1[i][j]) {
        printf("NO");
        return 0;
      }
  }
  printf("YES\n");
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) printf("%d ", temp[i][j]);
    printf("\n");
  }
  return 0;
}