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;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void XsliferX() {
  ios::sync_with_stdio(0);
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
}
long long f_p(long long x, int y) {
  if (y == 0)
    return 1;
  else if (y % 2 == 0)
    return f_p(x * x, y / 2);
  else
    return x * f_p(x * x, y / 2);
}
long long l_p(long long n) {
  n = n | (n >> 1);
  n = n | (n >> 2);
  n = n | (n >> 4);
  n = n | (n >> 8);
  return (n + 1) >> 1;
}
int dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
int arr[101][101], brr[101][101];
int main() {
  XsliferX();
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      brr[i][j] = 1;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> arr[i][j];
      if (arr[i][j] == 0) {
        for (int ii = 0; ii < n; ii++) {
          brr[ii][j] = 0;
        }
        for (int jj = 0; jj < m; jj++) {
          brr[i][jj] = 0;
        }
      }
    }
  }
  int a = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      a = 0;
      for (int ii = 0; ii < n; ii++) {
        a |= brr[ii][j];
      }
      for (int jj = 0; jj < m; jj++) {
        a |= brr[i][jj];
      }
      if (arr[i][j] != a) {
        cout << "NO\n";
        return 0;
      }
    }
  }
  cout << "YES\n";
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cout << brr[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}