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.*;
public class ORinMatrix {
    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 ans[][]=new int[m][n];
        int ad[][]=new int[m][n];
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++)
                a[i][j]=sc.nextInt();
        }
        int rows[]=new int[m];
        int cols[]=new int[n];
        int rowsd[]=new int[m];
        int colsd[]=new int[n];
        for(int i=0;i<m;i++){
            rows[i]=1;
            for(int j=0;j<n;j++)
                rows[i]&=a[i][j];
        }
        for(int j=0;j<n;j++){
            cols[j]=1;
            for(int i=0;i<m;i++)
                cols[j]&=a[i][j];
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++)
                ans[i][j]=rows[i]&cols[j];
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(ans[i][j]==1){
                    rowsd[i]=1;
                    break;
                }
            }
        }
        for(int j=0;j<n;j++){
            for(int i=0;i<m;i++){
                if(ans[i][j]==1){
                    colsd[j]=1;
                    break;
                }
            }
        }
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                ad[i][j]=rowsd[i]|colsd[j];
            }
        }
        
        boolean flag=true;
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(ad[i][j]!=a[i][j]){
                    flag=false;
                    break;
                }
            }
        }
        if(!flag)
            System.out.println("NO");
        else{
            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();
                }
            }
    }
}