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 scanner {

	public static void main(String[] args) {
		Scanner obj=new Scanner(System.in);
		int n=obj.nextInt();
		int m=obj.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++)
			{
				a[i][j]=1;
			}
		}
		
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				b[i][j]=obj.nextInt();
				
				if(b[i][j]==0)
				{
					for(int g=0;g<m;g++)
					{
						a[i][g]=0;
					}
					for(int g=0;g<n;g++)
					{
						a[g][j]=0;
					}
					
				}
			}
		}
		
		
		
		
		
		int flag=0;
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				if(b[i][j]==1)
				{	int u=0;
					for(int g=0;g<m;g++)
					{
						if(a[i][g]==1)
							u=1;
					}
					for(int g=0;g<n;g++)
					{
						if(a[g][j]==1)
							u=1;
					}
					if(u==0)
					{
						System.out.println("NO");
						flag=1;
						break;
					}
				}
				if(flag==1)
					break;
			}
			if(flag==1)
				break;
		}
		if(flag==0)
		{
			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();
			}
			
		}
		
			
	}

}