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 ProB {
	static int n,m,ju;
	static int[][] aa=new int[105][105];
	static int[][] bb=new int[105][105];

	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		n=in.nextInt();m=in.nextInt();
		for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
			aa[i][j]=in.nextInt();
		for(int i=1;i<=n;i++)
		{
			ju=1;
			for(int j=1;j<=m;j++)
				if(aa[i][j]==0) { ju=0;break;}
			aa[i][0]=ju;
		}
		for(int j=1;j<=m;j++)
		{
			ju=1;
			for(int i=1;i<=n;i++)
				if(aa[i][j]==0) { ju=0;break;}
			aa[0][j]=ju;
		}
		for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			if(aa[i][j]==0) continue;
			if(aa[i][0]==1&&aa[0][j]==1) bb[i][j]=1;
		}
		for(int i=1;i<=n;i++)
		{
			ju=0;
			for(int j=1;j<=m;j++)
				if(bb[i][j]==1) { ju=1;break;}
			bb[i][0]=ju;
		}
		for(int j=1;j<=m;j++)
		{
			ju=0;
			for(int i=1;i<=n;i++)
				if(bb[i][j]==1) { ju=1;break;}
			bb[0][j]=ju;
		}
		ju=1;
		o:
		for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++)
		{
			if(aa[i][j]==0) continue;
			if(bb[i][0]==0&&bb[0][j]==0) { ju=0;break o;}
		}
		System.out.println((ju==0)?"NO":"YES");
		if(ju==1)
		{
			for(int i=1;i<=n;i++)
			{
				for(int j=1;j<=m;j++)
					System.out.print(bb[i][j]+" ");
				System.out.println();
			}
		}
	}
}