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.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner t =new Scanner(System.in);
        int m=t.nextInt();
        int n=t.nextInt();
        boolean[][] b=new boolean[m][n];
        for (int i=0;i<m;i++){
            for (int j=0;j<n;j++){
                b[i][j]=t.nextInt()==0?false:true;
            }
        }
        
        StringBuilder str=new StringBuilder();
        str.append("YES\n");
        boolean[][] a=new boolean[m][n];
        boolean[][] c=new boolean[m][n];
        int cont=0;
        for (int i=0;i<m;i++){
            for (int j=0;j<n;j++){
                a[i][j]=true;
                for (int k=0;k<n;k++){
                    a[i][j]&=b[i][k];
                    if(!a[i][j]) break;
                }
                if(!a[i][j]){
                    if(j<n-1) str.append("0 ");
                    else str.append("0\n");
                    continue;
                }
                for (int k=0;k<m;k++){
                    a[i][j]&=b[k][j];
                    if(!a[i][j]) break;
                }
                if(a[i][j]){
                    for (int k=0;k<m;k++) c[k][j]=true;
                    for (int k=0;k<n;k++) c[i][k]=true;
                }
                if(j<n-1) str.append(a[i][j]?"1 ":"0 ");
                else str.append(a[i][j]?"1\n":"0\n");
            }
        }

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