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_matrix {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner in = new Scanner(System.in);
		int m = in.nextInt();
		int n = in.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++) {
				B[i][j] = in.nextInt();
			}
		}
		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++) {
				if (B[i][j] == 0) {
					for (int k = 0; k < n; k++) {
						A[i][k] = 0;
					}
					for (int l = 0; l < m; l++) {
						A[l][j] = 0;
					}
				}
			}
		}

		int counter = 0;
		boolean cont = true;
		for (int i = 0; i < m && cont; i++) {
			for (int j = 0; j < n; j++) {
				int test = 0;
				for (int k = 0; k < n; k++) {
					if (A[i][k] == 1) {
						test = 1;
						break;
					}
				}
				if (test == 0) {
					for (int l = 0; l < m; l++) {
						if (A[l][j] == 1) {
							test = 1;
							break;
						}
					}
				}
				if (B[i][j] != test) {
					cont = false;
					break;
				} else {
					counter++;
				}
			}
		}
		if (counter == n * m) {
			System.out.println("YES");
			for (int i = 0; i < m; i++) {
				for (int j = 0; j < n; j++) {
					System.out.print(A[i][j]);
					if(j<n-1){
						System.out.print(" ");
					}
				}
				System.out.println();
			}
		}
		else{
			System.out.println("NO");
		}
	}
}