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 maxn = 1024;
const long long INF = 1e15 + 7;
int n, m;
int a[maxn][maxn];
void check() {
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(1); j <= int(m); ++j) {
      if (!a[i][j]) continue;
      int flag1 = 1, flag2 = 1;
      for (int k = int(1); k <= int(n); ++k) flag1 &= a[k][j];
      for (int k = int(1); k <= int(m); ++k) flag2 &= a[i][k];
      if (!flag1 && !flag2) {
        printf("NO");
        exit(0);
      }
      bool checker = 0;
      if (flag1)
        for (int k = int(1); k <= int(n); ++k)
          checker |= (a[k][j - 1] & a[k][j + 1]);
      if (flag2)
        for (int k = int(1); k <= int(m); ++k)
          checker |= (a[i - 1][k] & a[i + 1][k]);
      if (!checker) {
        printf("NO");
        exit(0);
      }
    }
  }
  printf("YES\n");
}
int res[maxn][maxn];
int main() {
  scanf("%d %d", &n, &m);
  for (int i = int(1); i <= int(max(n, m)); ++i) a[i][0] = a[0][i] = 1;
  for (int i = int(1); i <= int(n); ++i) a[i][m + 1] = 1;
  for (int i = int(1); i <= int(m); ++i) a[n + 1][i] = 1;
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(1); j <= int(m); ++j) {
      scanf("%d", &a[i][j]);
    }
  }
  check();
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(1); j <= int(m); ++j) {
      if (!a[i][j]) continue;
      bool checker = 1;
      for (int k = int(1); k <= int(n); ++k) checker &= a[k][j];
      for (int k = int(1); k <= int(m); ++k) checker &= a[i][k];
      res[i][j] = checker;
    }
  }
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(1); j <= int(m); ++j) printf("%d ", res[i][j]);
    putchar('\n');
  }
  return 0;
}