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 a[101][101], b[101][101], t1[101], t2[101], hh[101], lh[101], i, j, k, m,
      n, c, he, le;
  ios::sync_with_stdio(false);
  c = 0;
  he = 0;
  le = 0;
  cin >> m >> n;
  memset(a, 0, sizeof(a));
  memset(t1, 0, sizeof(t1));
  memset(t2, 0, sizeof(t2));
  for (i = 0; i < m; i++)
    for (j = 0; j < n; j++) {
      cin >> a[i][j];
      t1[i] += a[i][j];
      t2[j] += a[i][j];
    }
  c = 0;
  for (i = 0; i < m; i++)
    for (j = 0; j < n; j++)
      if (a[i][j] == 1) {
        if ((t1[i] < n) && (t2[j] < m)) c = 1;
        if (c == 1) break;
      }
  if (c == 1)
    cout << "NO";
  else {
    he = 0;
    le = 0;
    for (i = 0; i < m; i++)
      if (t1[i] == n) {
        he++;
        hh[he] = i;
      }
    for (j = 0; j < n; j++)
      if (t2[j] == m) {
        le++;
        lh[le] = j;
      }
    for (i = 0; i < m; i++)
      for (j = 0; j < n; j++) b[i][j] = a[i][j];
    memset(a, 0, sizeof(a));
    for (i = 1; i <= he; i++)
      for (j = 1; j <= le; j++) a[hh[i]][lh[j]] = 2;
    for (i = 0; i < m; i++)
      for (j = 0; j < n; j++)
        if (a[i][j] == 2) {
          for (k = 0; k < m; k++)
            if (a[k][j] == 0) a[k][j] = 1;
          for (int l = 0; l < n; l++)
            if (a[i][l] == 0) a[i][l] = 1;
          a[i][j] = 1;
        }
    c = 0;
    for (i = 0; i < m; i++)
      for (j = 0; j < n; j++)
        if (b[i][j] != a[i][j]) c = 1;
    if (c == 0) {
      cout << "YES" << endl;
      memset(a, 0, sizeof(a));
      for (i = 1; i <= he; i++)
        for (j = 1; j <= le; j++) a[hh[i]][lh[j]] = 1;
      for (i = 0; i < m; i++) {
        for (j = 0; j < n; j++) cout << a[i][j] << " ";
        cout << endl;
      }
    } else
      cout << "NO";
  }
}