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 int binpow(long long int a, long long int b) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b = b >> 1;
  }
  return ans;
}
long long int binpowM(long long int a, long long int b,
                      long long int m = 1000000007) {
  if (a == 0) return 0;
  if (b == 0) return 1;
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a;
    ans %= m;
    a = a * a;
    a %= m;
    b = b >> 1;
  }
  return ans;
}
template <typename T>
T Min(T a, T b) {
  return a < b ? a : b;
}
template <typename T>
T Max(T a, T b) {
  return a > b ? a : b;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<vector<int> > a(n, vector<int>(m, 1));
  vector<vector<int> > c(n, vector<int>(m, 0));
  vector<int> p(n, 1), q(m, 1), x(n, 0), y(m, 0);
  auto b = a;
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (m); ++j) {
      cin >> b[i][j];
      p[i] &= b[i][j];
      q[j] &= b[i][j];
    }
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (m); ++j) {
      a[i][j] &= p[i] & q[j];
    }
  p = x, q = y;
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (m); ++j) {
      p[i] |= a[i][j];
      q[j] |= a[i][j];
    }
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (m); ++j) {
      c[i][j] |= p[i] | q[j];
    }
  bool ch = 1;
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (m); ++j)
      if (b[i][j] != c[i][j]) {
        ch = 0;
        break;
      }
  if (ch) {
    cout << "YES" << '\n';
    for (int i = 0; i < (n); ++i) {
      for (int j = 0; j < (m); ++j) cout << a[i][j] << " ";
      cout << '\n';
    }
  } else
    cout << "NO" << '\n';
  return 0;
}