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 in=new Scanner(System.in);
        int n=in.nextInt();
        int m=in.nextInt();
        int a[][]=new int[n][m];
        int b[][]=new int[n][m];
        int r[]=new int[n];
        int c[]=new int[m];
        int i,j,s,k;
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
             a[i][j]=in.nextInt();
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
             b[i][j]=1;
            }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(a[i][j]==0)
                {
                    for(k=0;k<m;k++)
                      b[i][k]=0;
                    for(k=0;k<n;k++)
                      b[k][j]=0;
                    }
                }
            }
        for(i=0;i<n;i++)
        {
            s=0;
            for(j=0;j<m;j++)
            {
                s=s|b[i][j];
            }
            r[i]=s;
        }
        for(i=0;i<m;i++)
        {
            s=0;
            for(j=0;j<n;j++)
            {
                s=s|b[j][i];
            }
            c[i]=s;
        }
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
            {
                if(a[i][j]!=(r[i]|c[j]))
                {
                    System.out.println("NO");
                    System.exit(0);
                }
            }
        }
        System.out.println("YES");
        for(i=0;i<n;i++)
        {
            for(j=0;j<m;j++)
             System.out.print(b[i][j]+" ");
            System.out.println();
        }
    }
}