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.Arrays;
import java.util.Scanner;
public class Watermelon {
public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
      int m=sc.nextInt();
      int n= sc.nextInt();
      int a[][]=new int [n][m];
      int b[][]=new int [n][m];
      int c=0,c1=0;
      
      for(int i=0; i<m;i++)
      {
          for(int j=0;j<n;j++)
          {
              b[j][i]=1;
          }
      }
    for(int i=0; i<m;i++)
      {
          for(int j=0;j<n;j++)
          {
              a[j][i]=sc.nextInt();
          }
      }
      
      
     for(int i=0; i<m;i++)
      {
          for(int j=0;j<n;j++)
          {
              if(a[j][i]==0)
              {
                 for(int k=0;k<n;k++)
                 {   
                     
                     b[k][i]=0;
                     
                 }
              for(int k=0;k<m;k++)
                 {   
                     
                     b[j][k]=0;
                     
                     
                 }
              
              }
        }
          
      }
     
     
     
     for(int i=0; i<m;i++)
      {
           
          for(int j=0;j<n;j++)
          {
              c=0;
                 for(int k=0;k<n;k++)
                 {   
                     
                     c=c|b[k][i];
                     
                 }
              for(int k=0;k<m;k++)
                 {   
                     
                    c=c|b[j][k];
                     
                     
                 }
              
              if(c!=a[j][i])
              {
                 c1=1;
                 break;
              }
        
          }
          
           if(c1==1)
           {
               System.out.println("NO");
               break;
           }
      }
      
     
     if(c1!=1)
     {
         
     System.out.println("YES");
     for(int i=0; i<m;i++)
      {
          for(int j=0;j<n;j++)
          {
              System.out.print(b[j][i]+" ");
          }
        System.out.println();
      }
     
     }
     
    
}

    
}