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;
inline void end(string s) {
  cout << s;
  exit(0);
}
inline void end(long long n) {
  cout << n;
  exit(0);
}
int mod(int a, int b) { return (((a % b) + b) % b); }
const int N = 105;
int r, c;
int A[N][N], B[N][N], C[N][N];
void erase(char t, int x, int v, int A[N][N]) {
  if (t == 'r') {
    for (int i = 0; i < N; ++i) {
      A[x][i] = v;
    }
  } else {
    for (int i = 0; i < N; ++i) {
      A[i][x] = v;
    }
  }
}
bool isSame(int A[N][N], int B[N][N]) {
  for (int i = 0; i < r; i++)
    for (int j = 0; j < c; j++)
      if (A[i][j] != B[i][j]) return false;
  return true;
}
int main(int argc, char const *argv[]) {
  cin >> r >> c;
  for (int i = 0; i < r; i++)
    for (int j = 0; j < c; j++) A[i][j] = 1;
  for (int i = 0; i < r; i++)
    for (int j = 0; j < c; j++) {
      cin >> B[i][j];
      if (!B[i][j]) {
        erase('r', i, 0, A);
        erase('c', j, 0, A);
      }
    }
  for (int i = 0; i < r; i++) {
    for (int j = 0; j < c; j++) {
      if (A[i][j]) {
        erase('r', i, 1, C);
        erase('c', j, 1, C);
      }
    }
  }
  if (isSame(B, C)) {
    cout << "YES" << endl;
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < c; j++) {
        cout << A[i][j] << " ";
      }
      cout << endl;
    }
  } else {
    cout << "NO";
  }
  return 0;
}