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.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Set;





public class Main {
 

  
    
    public static void main(String[] args) {
        
   
    
    
       
        Scanner sc=new Scanner(System.in);
    
        int n=sc.nextInt();
        int m=sc.nextInt();
        int[][] a=new int[n][m];
        int[] row=new int[n];
        int[] col=new int[m];
        boolean zero=true;
        for (int i = 0; i < n; i++) {
            
            for (int j = 0; j < m; j++) {
                
                a[i][j]=sc.nextInt();
                row[i]+=a[i][j];
                col[j]+=a[i][j];
                if(a[i][j]!=0)zero=false;
            }
           
        }
        boolean f=false;
        boolean ff=false;
        boolean ffc=false;
        for (int i = 0; i < n; i++) {
            if(row[i]==m&&!ff)ff=true;
            for (int j = 0; j < m; j++) {
                if(col[j]==n&&!ffc)ffc=true;
               if(a[i][j]==1)
               {
                   if(row[i]!=m&&col[j]!=n)
                   {f=true;
                   break;
                   }
               }
            }
            if(f)
           break;
        }
        
        if(f||(!ff||!ffc)&&!zero)
        {
            
            System.out.println("NO");
        }
        else
        {
            int num=0;
             System.out.println("YES");
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    if(row[i]==m&&col[j]==n)
                        a[i][j]=1;
                    else
                    { a[i][j]=0;
                       
                   
                   }
                     System.out.print((j==m-1)?a[i][j]+"":a[i][j]+" ");
               
                    
                  
                }
               System.out.println("");
            }
            
           
            
        }
      
}
}