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;
int N, M;
int A[100][100];
int B[100][100];
int C[100][100];
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) A[i][j] = 1;
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) {
      scanf("%d", B[i] + j);
      if (B[i][j]) continue;
      for (int k = 0; k < N; k++) A[k][j] = 0;
      for (int k = 0; k < M; k++) A[i][k] = 0;
    }
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) {
      for (int k = 0; k < N; k++) C[i][j] |= A[k][j];
      for (int k = 0; k < M; k++) C[i][j] |= A[i][k];
    }
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++)
      if (B[i][j] != C[i][j]) {
        printf("NO\n");
        return 0;
      }
  printf("YES\n");
  for (int i = 0; i < N; i++)
    for (int j = 0; j < M; j++) printf("%d%c", A[i][j], " \n"[j == M - 1]);
  return 0;
}