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.lang.reflect.Array;
import java.util.*;
public class cp {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int m,n;
        m=sc.nextInt();
        n=sc.nextInt();

        int a[][]=new int[m][n];

        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++)
                a[i][j]=sc.nextInt();


            HashMap<Integer,Integer>row=new HashMap<>();
            HashMap<Integer,Integer>col=new HashMap<>();

            for(int i=0;i<m;i++)
                for(int j=0;j<n;j++)
                {
                    if(a[i][j]==0) {
                        row.put(i, 1);
                        col.put(j, 1);
                    }
                }
            boolean f=true;
            for(int i=0;i<m;i++) {
                for (int j = 0; j < n; j++) {
                    if (a[i][j] == 1) {
                        if ((row.containsKey(i) && col.containsKey(j)) || row.size()==m || col.size()==n) {
                            f = false;
                            break;
                        }

                    }
                }
                if(!f)
                    break;
            }
            if(!f)
                System.out.println("NO");
            else{
                int ans[][]=new int[m][n];
                for(int k:row.keySet()){
                    for(int j=0;j<n;j++)
                        ans[k][j]=-1;
                }
                for(int k:col.keySet()){
                    for(int i=0;i<m;i++)
                        ans[i][k]=-1;
                }

                for(int i=0;i<m;i++)
                    for(int j=0;j<n;j++)
                        if(ans[i][j]==-1)
                            ans[i][j]=0;
                        else
                            ans[i][j]=1;
                System.out.println("YES");
                for(int i=0;i<m;i++) {
                    for (int j = 0; j < n; j++){
                        System.out.print(ans[i][j]+" ");
                    }
                    System.out.println();
                }
            }

    }
}