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 Main {
	static int m;
	static int n;
	public static void check(int i,int j,int arrayB[][],int arrayA[][]){
		int cnt=0;
		for(int k=0;k<m;k++){
			if(arrayB[k][j]==1)cnt++;
			//if(arrayB[i][k]==1)cnt++;
		}
		for(int k=0;k<n;k++)if(arrayB[i][k]==1)cnt++;
		if(cnt==m+n)arrayA[i][j]=1;
	   }
	   public static void put(int i,int j,int checkB[][]){
	   	for(int k=0;k<m;k++)
			checkB[k][j]=1;
			for(int k=0;k<n;k++)
			checkB[i][k]=1;
	   }
public static void main(String [] args){
Scanner in=new Scanner(System.in);
 m=in.nextInt();
 n=in.nextInt();
int arrayB[][]=new int[m][n];
int arrayA[][]=new int[m][n];
int checkB[][]=new int [m][n];
for(int i=0;i<m;i++)
	for(int j=0;j<n;j++)
		arrayB[i][j]=in.nextInt();
		
		for(int i=0;i<m;i++)
	    for(int j=0;j<n;j++)
	    if(arrayB[i][j]==1)
	    	check(i,j,arrayB,arrayA);	
	    for(int i=0;i<m;i++)
	    for(int j=0;j<n;j++)
	    	if(arrayA[i][j]==1)put(i,j,checkB);
	    	boolean flag=true;
	    for(int i=0;i<m;i++)
	    for(int j=0;j<n;j++)
	    	if(arrayB[i][j]!=checkB[i][j])flag=false;
	    	if(!flag)System.out.print("NO");
	    	else{
	    		System.out.println("YES");
	    		for(int i=0;i<m;i++){
	    			for(int j=0;j<n;j++){
	    			System.out.print(arrayA[i][j]+" ");
	    	}
	    	System.out.println();
	    }
	    	}
}
}