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.math.BigInteger;
import java.util.*;
public class ass {
        public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int m=in.nextInt();
        int n=in.nextInt();
        int sum1=0,sum2=0;
        int t[][]=new int[m][n];
        int u[][]=new int[m][n];
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
            u[i][j]=1;  
            }}
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                t[i][j]=in.nextInt();
            }}
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
            if(t[i][j]==1){
                for(int r=0;r<m;r++){
                if(t[r][j]==0){
                    for(int q=0;q<n;q++){
                        if(t[i][q]==0){
                            System.out.print("NO");
                        return; 
                        }}}}}}}
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
            if(t[i][j]==0){
                for(int r=0;r<m;r++){
                    u[r][j]=0;}
                for(int q=0;q<n;q++){
                    u[i][q]=0;
                }
            }}}
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(u[i][j]==0)
                    sum1++;
                if(t[i][j]==1)
                 sum2++ ;    
            }}
        if(sum1==m*n&&sum2>0){
            System.out.print("NO");
            return;
        }
        System.out.println("YES");
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                System.out.print(u[i][j]+" ");
            }
            System.out.println();   
        }
}
}