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 long long int MOD = 1e9 + 7;
bool IsPowerOfTwo(long long int x) { return (x != 0) && ((x & (x - 1)) == 0); }
long long int poer(long long int a, long long int b) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs *= a;
    a *= a;
    b /= 2;
  }
  return rs;
}
long long int mod_poer(long long int a, long long int b, long long int mo) {
  long long int rs = 1;
  while (b > 0) {
    if (b & 1) rs = (rs * a) % mo;
    a = (a * a) % mo;
    b /= 2;
  }
  return rs % mo;
}
long long int d, x, y;
void ext_euclid(long long int a, long long int b) {
  if (b == 0) {
    d = a;
    x = 1;
    y = 0;
  } else {
    ext_euclid(b, a % b);
    long long int temp = x;
    x = y;
    y = temp - (a / b) * y;
  }
}
long long int mod_inverse1(long long int a, long long int m) {
  ext_euclid(a, m);
  return (x % m + m) % m;
}
long long int mod_inverse2(long long int a, long long int m) {
  return mod_poer(a, m - 2, m);
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int main() {
  int m, n;
  cin >> m >> n;
  int b[m][n], i, j, a[m][n], k, c = 0, dd = 0, xx, yy;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      cin >> b[i][j];
    }
  }
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) a[i][j] = 1;
  }
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      if (b[i][j] == 0) {
        for (k = 0; k < n; k++) a[i][k] = 0;
        for (k = 0; k < m; k++) a[k][j] = 0;
      }
    }
  }
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) {
      dd = 0;
      if (b[i][j] == 1) {
        for (yy = 0; yy < n; yy++) {
          if (a[i][yy] == 0) dd++;
        }
        for (xx = 0; xx < m; xx++) {
          if (a[xx][j] == 0) dd++;
        }
        if (dd == (m + n)) {
          cout << "NO";
          return 0;
        }
      }
    }
  }
  cout << "YES" << endl;
  for (i = 0; i < m; i++) {
    for (j = 0; j < n; j++) cout << a[i][j] << " ";
    cout << endl;
  }
}