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.Scanner;


public class B {
    public static void main(String args[]){
        Scanner scan = new Scanner(System.in);
        int n,m;
        m=scan.nextInt();
        n=scan.nextInt();
        int B[][] = new int[m][n];
        for (int i=0;i<m;i++){
            for (int j=0;j<n;j++){
                B[i][j]=scan.nextInt();
            }
        }
        
        int A[][]=new int[m][n];
        for (int i=0;i<m;i++){
            for (int j=0;j<n;j++){
                A[i][j]=1;
            }
        }
            
        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 l=0;l<m;l++){
                        A[l][j]=0;
                    }
                }
            }
        }
        
        boolean flag=true;
        exit: for (int i=0;i<m;i++){
                for (int j=0;j<n;j++){
                if(B[i][j]==1){
                    flag=false;
                    for(int k=0;k<n;k++){
                        if(A[i][k]==1){flag=true;}
                    }
                    for(int l=0;l<m;l++){
                        if(A[l][j]==1){flag=true;}
                    }
                }
                if (B[i][j]!=0&!flag){System.out.println("NO"); break exit;}
        }
        
    }
        if(flag){
        System.out.println("YES");
        for (int z=0;z<m;z++){
            
            for (int x=0;x<n;x++){
                System.out.print(A[z][x]+" ");
            }
            System.out.println();
        }
        }

    }
}