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 sol{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		String[] t = sc.nextLine().split(" ");
		int n = Integer.parseInt(t[0]);
		int m = Integer.parseInt(t[1]);
		int[][] b = new int[n][m];
		int[][] a = new int[n][m];
		for(int i = 0;i< n;i++){
		    for(int j=0;j< m;j++){
		        a[i][j] = 1;
		    }
		}
		for(int i = 0;i< n;i++){
		    for(int j=0;j< m;j++){
		        int x = sc.nextInt();
		        b[i][j] = x;
		        if(x == 0){
		            for(int k = 0;k<m;k++){
		                a[i][k] = 0;
		            }
		            for(int k = 0;k<n;k++){
		                a[k][j] = 0;
		            }
		        }
		    }
		}
		boolean flag = true;
		for(int i = 0; flag && i< n;i++){
		    for(int j=0;flag && j< m;j++){
		        
		        if(b[i][j] == 1){
		            int k;
		            int l;
		            for( k = 0;k<m;k++){
		                if(a[i][k] == 1){
		                    break;
		                }
		            }
		            for( l = 0;l<n;l++){
		                if(a[l][j] == 1){
		                    break;
		                }
		            }
		            if(k == m && l == n){
		                flag = false;
		            }
		            
		        }
		    }
		}
		if(flag){
		    System.out.println("YES");
		    for(int i=0; i<n; i++){
                System.out.print(a[i][0]);
                
                for(int j=1; j<m; j++)
                    {System.out.print(" ");
                    System.out.print(a[i][j]);}
                 System.out.println();
            } 
		}
		else{
		    System.out.print("NO");
		}
		
		
	}
}