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 abc
{
    public static void main(String ar[])
    {
        Scanner sc = new Scanner(System.in);
        int m=sc.nextInt(); int n=sc.nextInt(); boolean q=true;
        int c[][]=new int[m][n];
        int b[][]=new int[m][n];
        int a[][]=new int[m][n];
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                b[i][j]=sc.nextInt();
                a[i][j]=1;
                c[i][j]=0;
            }    
        }
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(b[i][j]==0)
                {
                    for(int k=0;k<n;k++)
                    a[i][k]=0;
                    for(int k=0;k<m;k++)
                    a[k][j]=0;
                }
            }
        }
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                for(int k=0;k<n;k++)
                c[i][j]=c[i][j]|a[i][k];
                for(int k=0;k<m;k++)
                c[i][j]=c[i][j]|a[k][j];
            }
        }
        a1:for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(c[i][j]!=b[i][j])
                {q=false; break a1;}
            }
        }
        if(q)
        {
            System.out.println("YES");
            for(int i=0;i<m;i++)
            {
                for(int j=0;j<n;j++)
                System.out.print(a[i][j]+" ");
                System.out.println();
            }
        }
        else
        System.out.println("NO");
    }
}