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.* ;
import java.io.*;



public class main {
static int sol [] [] ;
    
    
static void fillit(int r , int c , int x , int sol [][]){
for(int i = 0 ; i < sol.length ; i++)
    sol[i][c] = x ;
for(int i = 0 ; i < sol[r].length ; i++)
    sol[r][i] = x ;


}
static void print2D(int n , int m , int ar [][] ){
StringBuilder ans = new StringBuilder() ;
for(int i = 0 ; i < n ; i++)
{for(int j = 0 ; j < m ; j++)
    ans.append(ar[i][j] + " ") ;
    ans.append('\n') ;
    
    }       

    System.out.println(ans);
}
static boolean cheack(int ar[] [] ,int ar2 [] []){
for(int i = 0 ; i < ar.length ; i++)
    for(int j = 0 ; j < ar[i].length ; j++)
        if( ar[i][j] != ar2[i][j])
            return false ;
return true ;

}

    public static void main(String[] args) throws Exception{
     BufferedReader bf = new BufferedReader(new InputStreamReader(System.in)) ;
     StringTokenizer str = new StringTokenizer(bf.readLine()) ;
     int n = Integer.parseInt(str.nextToken()) ;
     int m = Integer.parseInt(str.nextToken()) ;  
      sol = new int[n][m] ;
     for(int i = 0 ; i < n ; i++)
         Arrays.fill(sol[i], 1);
       
    
     
     
     int ar [][] = new int[n][m] ;
     for(int i = 0 ; i < n ; i++){
         str = new StringTokenizer(bf.readLine()) ;
         for(int j = 0 ; j < m ;j++)
         { ar[i][j] = Integer.parseInt(str.nextToken()) ;
             if(ar[i][j] == 0)
                 fillit(i, j,0,sol) ;
                 
                 }
         
       
         
         
                     }
     
       int rr [] [] = new int [n][m] ;
       for(int i = 0 ; i < n ; i++)
           for(int j = 0 ; j < m ;j++)
               if(sol[i][j] == 1)
                   fillit(i, j, 1, rr) ;
       
        
      if(cheack(ar, rr)) 
      {   System.out.println("YES");
          print2D(n, m, sol);}
      else 
            System.out.println("NO");
          }}