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>
const int maxN = (int)3e5 + 100;
const int MOD = (int)100000000;
const unsigned int Modu = 1000000007;
int a[maxN];
using namespace std;
template <class ForwardIterator>
ForwardIterator next(
    ForwardIterator it,
    typename iterator_traits<ForwardIterator>::difference_type n = 1);
bool IsPrime(int number) {
  int root = sqrt(number);
  for (int i = 2; i <= root; i++) {
    if (number % i == 0) return false;
  }
  return true;
}
int countDivisors(int n) {
  int cnt = 0;
  for (int i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      if (n / i == i)
        cnt++;
      else
        cnt = cnt + 2;
    }
  }
  return cnt;
}
long int binpow(long int a, long int b) {
  long int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = ((res) * (a));
    }
    a = (a * a);
    b >>= 1;
  }
  return res;
}
void solve() {
  int m, n;
  cin >> m >> n;
  int arr[m][n];
  for (int i = 0; i < (m); ++i) {
    for (int j = 0; j < (n); ++j) {
      arr[i][j] = 1;
    }
  }
  int cnt1 = 0, cnt2 = 0;
  for (int i = 0; i < (m); ++i) {
    for (int j = 0; j < (n); ++j) {
      int x;
      cin >> x;
      if (x == 0) {
        for (int k = 0; k < (n); ++k) {
          arr[i][k] = 0;
        }
        for (int k = 0; k < (m); ++k) {
          arr[k][j] = 0;
        }
      } else {
        cnt1++;
      }
    }
  }
  for (int i = 0; i < (m); ++i) {
    for (int j = 0; j < (n); ++j) {
      int flag = 0;
      for (int k = 0; k < (n); ++k) {
        if (arr[i][k] == 1) {
          flag = 1;
          break;
        }
      }
      if (flag == 0) {
        for (int k = 0; k < (m); ++k) {
          if (arr[k][j] == 1) {
            flag = 1;
            break;
          }
        }
      }
      if (flag == 1) {
        cnt2++;
      }
    }
  }
  if (cnt1 == cnt2) {
    cout << "YES" << endl;
    ;
    for (int i = 0; i < (m); ++i) {
      for (int j = 0; j < (n); ++j) {
        cout << arr[i][j] << " ";
        ;
      }
      cout << endl;
    }
  } else {
    cout << "NO" << endl;
    ;
  }
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
}