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 Solution {
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		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++){
		        B[i][j] = sc.nextInt();
		        A[i][j] = B[i][j];
		    }
		}
		
		for(int i=0;i<n;i++){
		    for(int j=0;j<m;j++){
		        if(B[i][j] == 0){
		            for(int k=0;k<m;k++){
		                A[i][k] = 0;
		            }
		            for(int k=0;k<n;k++){
		                A[k][j] = 0;
		            }
		        }
		    }
		}
		int C[][] = new int[n][m];
		for(int i=0;i<n;i++){
		    for(int j=0;j<m;j++){
		        if(A[i][j] == 1){
		            for(int k=0;k<m;k++){
		                C[i][k] = 1;
		            }
		            for(int k=0;k<n;k++){
		                C[k][j] = 1;
		            }
		        }
		    }
		}
		boolean flag = true;
		for(int i=0;i<n;i++){
		    for(int j=0;j<m;j++){
		        if(B[i][j] != C[i][j]){
		            flag = false;
		            break;
		        }
		    }
		}
		if(flag){
		    System.out.println("YES");
		    for(int i=0;i<n;i++){
    		    for(int j=0;j<m;j++){
    		        System.out.print(A[i][j]+" ");
    		    }
    		    System.out.println();
    		}
		}else{
		    System.out.println("NO");
		}
	}
}