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 Main {
    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int row = sc.nextInt();
    int column = sc.nextInt();
    int[][] B = new int[row][column];
    int[][] A = new int[row][column];
    for(int i=0;i<row;i++){
        for(int j=0;j<column;j++){
            B[i][j] = sc.nextInt();
            A[i][j]=1;
        }
    }
    for(int i=0;i<row;i++){
        for(int j=0;j<column;j++){
            if(B[i][j]==0){
                for(int m=0;m<column;m++){
                    A[i][m]=0;
                }
                for(int m=0;m<row;m++){
                    A[m][j]=0;
                }
            }
        }
    }
    for(int i=0;i<row;i++){
        for(int j=0;j<column;j++){
            if(B[i][j]==1){
                boolean flag=false;
                 for(int m=0;m<column;m++){
                    if(A[i][m]==1) flag=true;
                 }
                 for(int m=0;m<row;m++){
                    if( A[m][j]==1) flag=true;
                 }
                 if(!flag){
                     System.out.println("NO");
                     System.exit(0);
                 }
            }
        }
    }
    System.out.println("YES");
    for(int i=0;i<row;i++){
        for(int j=0;j<column;j++){
            System.out.print(A[i][j]+" ");
        }
        System.out.println("");
    }
    }
}