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 JAVA):

Solution:
import java.util.*;

public class B {

      public static void main (String[] args) {
            
            Scanner S = new Scanner(System.in);
            int m = S.nextInt(), n = S.nextInt();
            int[][] A = new int [m][n], B = new int[m][n];
            
            for (int i = 0 ; i < m ; i++)
                  Arrays.fill(A[i], 1);
            
            for (int i = 0 ; i < m ; i++)
                  for (int j = 0 ; j < n ; j++)
                        if ((B[i][j] = S.nextInt()) == 0) {
                              for (int k = 0 ; k < m ; k++)
                                    A[k][j] = 0;
                              for (int k = 0 ; k < n ; k++)
                                    A[i][k] = 0;
                        }
            
            boolean valid = true;
            for (int i = 0 ; i < m && valid ; i++)
                  for (int j = 0 ; j < n && valid ; j++)
                        if (B[i][j] == 1) {
                              boolean one = false;
                              for (int k = 0 ; k < m && !one ; k++)
                                    if (A[k][j] == 1)
                                          one = true;
                              for (int k = 0 ; k < n && !one ; k++)
                                    if (A[i][k] == 1)
                                          one = true;
                              if (!one)
                                    valid = false;
                        }
            
            if (valid) {
                  System.out.println("YES");
                  for (int i = 0 ; i < m && valid ; i++) {
                        System.out.print(A[i][0]);
                        for (int j = 1 ; j < n && valid ; j++)
                              System.out.print(" " + A[i][j]);
                        System.out.println();
                  }
            } else
                  System.out.println("NO");
            
      }

}