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 asddfa {
	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];
		for (int k = 0; k < n; k++) {
			for (int j = 0; j < m; j++) {
				B[k][j] = sc.nextInt() == 1;
			}
		}
		int[][] A = new int[n][m];
		for (int k = 0; k < n; k++) {
			for (int j = 0; j < m; j++) {
				A[k][j] = 0;
			}
		}
		for (int k = 0; k < n; k++) {
			for (int j = 0; j < m; j++) {
				if (!B[k][j]) {
					for (int i = 0; i < n; i++) {
						A[i][j] = 1;
					}
					for (int i = 0; i < m; i++) {
						A[k][i] = 1;
					}
				}
			}
		}
		boolean[][] Bp = new boolean[n][m];
		boolean good = true;
		for (int k = 0; good && k < n; k++) {
			for (int j = 0; good && j < m; j++) {
				for (int i = 0; !Bp[k][j] && i < n; i++) {
					Bp[k][j] = (A[i][j] == 0);
				}
				for (int i = 0; !Bp[k][j] && i < m; i++) {
					Bp[k][j] = (A[k][i] == 0);
				}
				good = (Bp[k][j] == B[k][j]);
			}
		}
		if (good) {
			System.out.println("YES");
			for (int k = 0; k < n; k++) {
				for (int j = 0; j < m; j++) {
					if (j!=0)
						System.out.print(" ");
					System.out.print(A[k][j]^1);
				}
				System.out.println();
			}
		} else {
			System.out.println("NO");
		}
	}

	public static int gcd(int a, int b) {
		return (b == 0) ? a : gcd(b, a % b);
	}
}