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.Scanner;
import java.util.Stack;
    
    public class Solution {
       
    public static void main(String[]args){
        Scanner s=new Scanner(System.in);
        int m=s.nextInt();
        int n=s.nextInt();
        int[][] a=new int[m][n];
        int[][] b=new int[m][n];    
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                a[i][j]=1;
            }
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                b[i][j]=s.nextInt();
            }
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(b[i][j]==0)
                    fun(a,i,j,m,n);
            }
        }
        int[][] q=new int[m][n];
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                q[i][j]=a[i][j];
            }
            
        }
        
        
     int c=d(a,b,m,n);
     if(c==0){
         System.out.println("NO");
     
     }
     else{
         System.out.println("YES");
         for(int i=0;i<m;i++){
                for(int j=0;j<n;j++){
                    System.out.print(q[i][j]+" ");
                }
                System.out.println();
            }
     }
    }
    public static void fun(int a[][],int i,int j,int m,int n){
        for(int k=0;k<n;k++)
            a[i][k]=0;
        for(int k=0;k<m;k++)
            a[k][j]=0;
        
    }
    public static int d(int a[][],int b[][],int m,int n){
        int r=0;
        int c=0;
        
        for(int k=0;k<n;k++){
            if(a[0][k]==1)
            r++;
        }
        for(int k=0;k<m;k++){
            if(a[k][0]==1)
                c++;
        }
        
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                if(a[i][j]==1){
                    a[0][j]=1;
                    a[i][0]=1;
                }
            }
        }
        for(int i=1;i<m;i++){
            for(int j=1;j<n;j++){
                if(a[0][j]==1||a[i][0]==1){
                    a[i][j]=1;
                    
                }
            }
        }
        if(r>0){
        for(int k=0;k<n;k++)
            a[0][k]=1;
            
        }
        if(c>0){
        for(int k=0;k<m;k++)
            a[k][0]=1;
                
        }
        
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(a[i][j]!=b[i][j])
                    return 0;
            }
        }
        return 1;
    }
    }