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.*;
import java.io.*;
import java.lang.*;
public class Main{
    public static void main(String args[] ) throws Exception {
        Scanner scan =new Scanner(System.in);
        int m = scan.nextInt();
        int n = scan.nextInt();
        int[][] b = new int[m+1][n+1];
        int[][] a = new int[m+1][n+1];
        for(int i=1 ; i<m+1 ; i++){
            for(int j=1 ; j<n+1 ; j++){
                b[i][j]=scan.nextInt();
                a[i][j]=1;
            }
        }
        for(int i=1 ; i<m+1 ; i++){
            for(int j=1 ; j<n+1 ; j++){
                if(b[i][j]== 0){
                    for(int x=1 ; x<=n ;x++ ){
                        a[i][x]=0;
                    }
                     for(int y=1 ; y<=m ;y++ ){
                        a[y][j]=0;
                    }
                }
            }
        }
        String d ="YES";
        for(int i=1 ; i<m+1 ; i++){
            for(int j=1 ; j<n+1 ; j++){
                if (b[i][j]==1){
                    int r=0;
                    for(int x=1 ; x<=n ;x++ ){
                        r+=a[i][x];
                    }
                    for(int y=1 ; y<=m ;y++ ){
                        r+=a[y][j];
                    }
                    if(r == 0){
                        d="NO";
                    }
                }
               
            }
        }
        System.out.println(d);
        if(d.equals("YES")){
            for(int i=1 ; i<m+1 ; i++){
                for(int j=1 ; j<n+1 ; j++){
                    System.out.print(a[i][j]+" ");
             }
             System.out.println();
            }
        }
    }
}