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.io.*;
import java.util.*;
import java.math.*;
import java.text.*;
public class Main
{
public static void main(String[] args)throws java.lang.Exception
{
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
PrintWriter out=new PrintWriter(System.out);
int n,m,i,j,v,k;
boolean found,foundr0,foundc0;


StringTokenizer st1=new StringTokenizer(in.readLine());

m=Integer.parseInt(st1.nextToken());
n=Integer.parseInt(st1.nextToken());

int[][] a=new int[m][n];
int[][] b=new int[m][n];
int[][] c=new int[m][n];

for(i=0;i<m;i++)
{
StringTokenizer st2=new StringTokenizer(in.readLine());

for(j=0;j<n;j++)
	{
	a[i][j]=Integer.parseInt(st2.nextToken());
	}
}

found=false;

for(i=0;i<m;i++)
{
for(j=0;j<n;j++)
	{
	v=a[i][j];

	if(v==1)
	{

	foundr0=false;
	foundc0=false;

	for(k=0;k<m;k++)
		{
		if(k!=i)
			{
			if(a[k][j]==0)
				foundr0=true;
			}
		}
	
	for(k=0;k<n;k++)
		{
		if(k!=j)
			{
			if(a[i][k]==0)
				foundc0=true;
			}
		}
	
	if(foundr0 && foundc0)
		{
		found=true;
		}

	else
		{
		if(!foundr0 && !foundc0)
			b[i][j]=1;
		else
			b[i][j]=0;
		}

	}
	else
	{
	b[i][j]=0;
	}
	
	}
}




for(i=0;i<m;i++)
	{	
	for(j=0;j<n;j++)
		{
		for(k=0;k<m;k++)
			{
			if(b[k][j]==1)
				c[i][j]=1;
			}
	
		for(k=0;k<n;k++)
			{
			if(b[i][k]==1)
				c[i][j]=1;
			}		

		}
	}


/*
for(i=0;i<m;i++)
		{
		out.print(c[i][0]);
		
		for(j=1;j<n;j++)
			{
			out.print(" "+c[i][j]);
			}
		out.println();
		}
*/


for(i=0;i<m;i++)
	{	
	for(j=0;j<n;j++)
		{
		if(c[i][j]!=a[i][j])
			found=true;
		}
	}




if(found)
	{
	out.println("NO");
	}
else
	{
	out.println("YES");

	for(i=0;i<m;i++)
		{
		out.print(b[i][0]);
		
		for(j=1;j<n;j++)
			{
			out.print(" "+b[i][j]);
			}

		out.println();
		}
	}


out.flush();
out.close();
}
}