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.Arrays;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();

		boolean[][] b = new boolean[n][m];
		boolean[][] a = new boolean[n][m];
		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				b[i][j] = sc.next().equals("1");

		for (int i = 0; i < n; i++)
			for (int j = 0; j < m; j++)
				a[i][j] = test(b, i, j, n, m);

		for(int i = 0; i < n; i++)
			for(int j = 0; j < m; j++){
				if(b[i][j] != test2(a, i, j, n, m)){
					System.out.println("NO");
					System.exit(0);
				}
			}
		
		System.out.println("YES");
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++)
				System.out.print((a[i][j] ? "1" : "0") + " ");
			System.out.println();
		}
		
	}

	private static boolean test2(boolean[][] a, int x, int y, int n, int m) {
		boolean res = false;
		for(int i = 0; i < n; i++)
			res |= a[i][y];
		for(int i = 0; i < m; i++)
			res |= a[x][i];
		return res;
	}

	private static boolean test(boolean[][] b, int x, int y, int n, int m) {
		boolean res = b[x][y];
		for (int i = 0; i < n; i++)
			res &= b[i][y];
		for (int i = 0; i < m; i++)
			res &= b[x][i];
		return res;
	}
}