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 helloWorld 
{
	public static void main(String[] args) 
	{		
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		int[][] A = new int[n][m];
		int[][] B = new int[n][m];
		String ans = "YES";
		
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++) {
				B[i][j] = in.nextInt();
				A[i][j] = 1;
			}
		
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++)
				if(B[i][j] == 0) {
					for(int k = 0; k < n; k++)
						A[k][j] = 0;
					for(int k = 0; k < m; k++)
						A[i][k] = 0;
				}
		
		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++)
				if(B[i][j] == 1) {
					int a = 0;
					for(int k = 0; k < n; k++)
						a += A[k][j];
					for(int k = 0; k < m; k++)
						a += A[i][k];
					if(a == 0)
						ans = "NO";
				}
		
		System.out.println(ans);
		if(ans.equals("YES"))
			for(int i = 0; i < n; i++) {
				for(int j = 0; j < m; j++)
					System.out.printf(A[i][j] + " ");
				System.out.println();
			}
		
		//in.close();
	}
}