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:
//package feb19;

import java.util.Arrays;
import java.util.Scanner;

public class B486 {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		int rows=scn.nextInt();
		int cols=scn.nextInt();
		int[][] arr=new int[rows][cols];
		for(int i=0; i<rows; i++)
		{
			for(int j=0; j<cols; j++)
			{
				arr[i][j]=scn.nextInt();
			}
		}
		int[][] find = new int[rows][cols];
		/*Here is the catch. Lets all elements of required matrix be 1. I tried with 0 but it was
		 * tough there*/
		for(int i=0; i<rows; i++)
		{
			for(int j=0; j<cols; j++)
			{
				find[i][j]=1;
			}
		}
		for(int i=0; i<rows; i++)
		{
			for(int j=0; j<cols; j++)
			{
				if(arr[i][j]==0)
				{
					for(int k=0; k<rows; k++)
					{
						find[k][j]=0;
					}
					for(int k=0; k<cols; k++)
					{
						find[i][k]=0;
					}
				}
			}
		}
		
		int isprint=0;
		int ans=-1;
		for(int i=0; i<rows; i++)
		{
			for(int k=0; k<cols; k++)
			{
				ans=find[i][k];
				for(int k1=0; k1<cols; k1++)
				{
				ans=(ans| find[i][k1]);
				}
				for(int k1=0; k1<rows; k1++)
				{
				ans=(ans | find[k1][k]);
				}
				//System.out.println("I:"+i+"K:"+k+"Ans:"+ans);
					if(ans!=arr[i][k])
					{
						isprint=1;
						break;
					}
			}
			
		//	System.out.println();
		}
		if(isprint==0)
		{
			System.out.println("YES");
			for(int i=0; i<rows; i++)
			{
				for(int j=0; j<cols; j++)
				{
					System.out.print(find[i][j]+" ");
				}
				System.out.println();
			}
		}
		else
		{
			System.out.println("NO");
		}

	}

}