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 main2 {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int m = sc.nextInt();
		int n = sc.nextInt();
		int[][] b=  new int[m][n];
		int[][] a=  new int[m][n];
		for (int i = 0; i < b.length; i++) {
			for (int j = 0; j < b[0].length; j++) {
				b[i][j] = sc.nextInt();
				a[i][j] = 1;
			}
		}
		
		String yes = "YES";
		String no = "NO";
		
		for (int i = 0; i < b.length; i++) {
			for (int j = 0; j < b[0].length; j++) {
				if (b[i][j] == 0) {
					for (int j2 = 0; j2 < a[0].length; j2++) {
						a[i][j2] = 0;
					}
					
					for (int j2 = 0; j2 < a.length; j2++) {
						a[j2][j] = 0;
					}
				}
			}
		}
		
		for (int i = 0; i < b.length; i++) {
			for (int j = 0; j < b[0].length; j++) {
				
				int c = 0;
				for (int j2 = 0; j2 < a[0].length; j2++) {
					c+= a[i][j2];
				}
				
				for (int j2 = 0; j2 < a.length; j2++) {
					c+=a[j2][j] ;
				}
				
				if (c > 0) {
					c = 1;
				}
				
				if (c != b[i][j]) {
					System.out.println(no);
					return;
				}
			}
		}
		
		System.out.println(yes);
		for (int i = 0; i < a.length; i++) {
			StringBuilder sb = new StringBuilder();
			for (int j = 0; j < a[0].length; j++) {
				if (j < a[0].length - 1) {
					sb.append(a[i][j] + " ");
				} else {
					sb.append(a[i][j]);
				}
			}
			
			System.out.println(sb.toString());
		}
	}
}