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.Scanner;

public class OR {

	private static Scanner input;

	public static void main(String[] args) {
		input = new Scanner(System.in);
		int m = input.nextInt();
		int n = input.nextInt();
		int[][] B = new int[m][n];
		int[][] A = new int[m][n];
		
		
		for (int i = 0 ; i < m ; i++ ) {
			for ( int j = 0 ; j < n ; j++ ) {
				A[i][j] = 1 ;
			}
		}
		
		
		for (int i = 0 ; i < m ; i++ ) {
			for ( int j = 0 ; j < n ; j++ ) {
				B[i][j] = input.nextInt();
				if ( B[i][j] == 0 ) {
					for (int k = 0 ; k < n ; k++ ) {
						A[i][k] = 0 ;
					}
					for (int g = 0 ; g < m ; g++ ) {
						A[g][j] = 0 ;
					}
				}
			}
		}
		
		int flag = 0 ;
		int found = 0 ;
		for (int i = 0 ; i < m ; i++ ) {
			for ( int j = 0 ; j < n ; j++ ) {
				if (B[i][j] == 1) {
					found = 0 ;
					for (int k = 0 ; k < n ; k++ ) {
						if (A[i][k] == 1) {
							found = 1 ;
							break;
						}
					} 
					if (found == 0) {
						for (int g = 0 ; g < m ; g++ ) {
							if (A[g][j] == 1) {
								found = 1 ;
								break;
							}
						}
					}
				}
				if ( (found == 0) && (B[i][j] == 1) ) {
					flag = 1 ;
					break;
				}
			}
			if (flag == 1) {
				break;
			}
		}
		
		
		
		if (flag == 1) {
			System.out.println("NO");
		} else {
			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();
			}
		}
	}
}