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.util.Arrays;

public class orMatrix
{
    public static void main(String [] args)
    {
        Scanner in=new Scanner(System.in);
        int m=in.nextInt();
        int n=in.nextInt();
       int[][] a=new int[108][108];
        
        int[] rowSum=new int[108];
        Arrays.fill(rowSum,0);
        
        int[] columnSum=new int[108];
        Arrays.fill(columnSum, 0);
        int allSum=0;
        
        for (int i=0;i<m;i++)
            for (int j=0;j<n;j++)
            {
                a[i][j]=in.nextInt();
                rowSum[i]+=a[i][j];columnSum[j]+=a[i][j];allSum+=a[i][j];
            }
        
        
        
        ///condition1:check each element;
        // if this elemen equal with 1,least rowsum or columnsum reach max value 
        boolean condition1=true;
        for (int i=0;i<m;i++)
            for (int j=0;j<n;j++)
                if (a[i][j]==1)
                    if (rowSum[i]!=n && columnSum[j]!=m) condition1=false;
        
        ///condition 2:check :at least element has both rowsum and columnsum that reach the max value 
        ////or all element are equal with 0....
        boolean condition2 =false;
        for (int i=0;i<m;i++)
            for (int j=0;j<n;j++) 
                if (a[i][j]==1)
                    if (rowSum[i]==n && columnSum[j]==m) condition2=true;
        if (allSum==0) condition2 =true;
       
        
        ///exist result when both all condition are true ;
        if ((condition1 && condition2) ==false) System.out.println("NO");
        else 
        {
            System.out.println("YES");
            for (int i=0;i<m;i++)
            {
                for (int j=0;j<n;j++) 
                    if (rowSum[i]==n && columnSum[j]==m) System.out.print(1+" ");
                            else System.out.print(0+" ");
                System.out.println();
            }
        }
    }
}