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 ORinMatrix
{
	public static void main(String[] args)
	{
		Scanner sc=new Scanner(System.in);
		int m=sc.nextInt();
		int n=sc.nextInt();
		int a[][]=new int[m][n];
		int b[][]=new int[m][n];
		int c[][]=new int[m][n];
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				b[i][j]=sc.nextInt();
				a[i][j]=1;
			}
		}
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(b[i][j]==0)
				{
					for(int k=0;k<n;k++)
						a[i][k]=0;
					for(int k=0;k<m;k++)
						a[k][j]=0;
				}
			}
		}
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				//int flag=0;
				for(int k=0;k<n;k++)	
				{
					if(a[i][k]==1)
					{
						c[i][j]=1;
						//flag=1;
						//break;
						
					}
				}
				//if(flag==0)
				//{
					for(int k=0;k<m;k++)	
					{	
						if(a[k][j]==1)
						{
							c[i][j]=1;
							break;
						}							
					}	
				//}
					
				
			}
		}
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				if(b[i][j]!=c[i][j])
				{
					System.out.println("NO");
					return;
				}
			}			
		}
		System.out.println("YES");
		for(int i=0;i<m;i++)
		{
			for(int j=0;j<n;j++)
			{
				System.out.print(a[i][j]+" ");
			}			
			System.out.println();
		}
	}
}