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 Problem486B {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int n = sc.nextInt();
        int[][] A = new int[m][n];
        int[][] B = new int[m][n];
        int[][] C = new int[m][n];

        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                B[i][j] = sc.nextInt();
                A[i][j] = 1;
                C[i][j] = 0;
            }
        }
        sc.close();
        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 k = 0; k < m; ++k)
    			{
    				A[k][j] = 0;
    			}

    		}
    	}
    }
    
    for (int i = 0; i < m; ++i)
    {
    	for (int j = 0; j < n; ++j)
    	{
    		if (A[i][j] == 1)
    		{
    			for (int k = 0; k < n; ++k)
    			{
    				C[i][k] = 1;
    			}
    			for (int k = 0; k < m; ++k)
    			{
    				C[k][j] = 1;
    			}

    		}
    	}
    }
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                if(C[i][j] != B[i][j]){
                    System.out.println("NO");
                    return;
                }
            }
        }
        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();
        }
    }
    
}