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 M {
    
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int m = scan.nextInt();
        int n = scan.nextInt();
        int a[][] = new int[m][n];
        int s[][] = new int[m][n];
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                a[i][j] = scan.nextInt();
                s[i][j] = 1;
            }
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if(a[i][j]==0) {
                    for (int j2 = 0; j2 < n; j2++) {
                        s[i][j2] = 0;
                    }
                    for (int j2 = 0; j2 < m; j2++) {
                        s[j2][j] = 0;
                    }
                }
            }
        }
        boolean flag = true;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                boolean f = false;
                if(a[i][j]==1) {
                    for (int j2 = 0; j2 < n; j2++) {
                        if(s[i][j2] == 1) {
                            f=true;
                            break;
                        }
                    }
                    if(!f) {
                        for (int j2 = 0; j2 < m; j2++) {
                            if(s[j2][j] == 1) {
                                f=true;
                                break;
                            }
                            
                        }
                    }
                    if(!f) {
                        flag = false;
                        break;
                    }
                }
            }
            if(!flag) {
                break;
            }
        }
        if(flag) {
            System.out.println("YES");
            for (int i = 0; i < m; i++) {
                for (int j = 0; j < n; j++) {
                    System.out.print(s[i][j]);
                    if(j != n-1)
                        System.out.print(" ");
                }
                System.out.println();
            }
        }
        else
            System.out.println("NO");
        
    }

}