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 Main {
    static void print(int [][]a){
        int m= a.length;
        int n= a[0].length;
        for (int i = 0; i <m ; i++) {
            for (int j = 0; j <n ; j++) {
                System.out.print( a[i][j]+" ");
            }
            System.out.println();
        }
    }


    public static void main(String[] args) {
        // write your code here
        Scanner sc=new Scanner(System.in);
        int m=sc.nextInt();
        int n =sc.nextInt();
        int[] sumH=new int[m];
        int[] sumV=new int[n];
        int hor=0,ver=0;
        int tot=0;

        for (int i = 0; i <m ; i++) {
            for (int j = 0; j <n ; j++) {
                int one=sc.nextInt();
                tot+=one;
                sumH[i]+=one;
                sumV[j]+=one;
                if(sumH[i]==n)hor++;
                if(sumV[j]==m)ver++;
            }
        }
        int [][] la=new int[m][n];
        if(hor>0&&ver>0){


            for (int i = 0; i <m ; i++) {
                for (int j = 0; j <n ; j++) {
                    if(sumH[i]==n&&sumV[j]==m)la[i][j]=1;
                    if(sumH[i]<n&&sumV[j]<m && (sumH[i]!=ver||sumV[j]!=hor)){
                        System.out.println("NO");
                        return;
                    }
                }
            }
            System.out.println("YES");
            print(la);
        }
        else {
            if(tot==0)
            { System.out.println("YES");
                print(la);}
            else
                System.out.println("NO");}
    }
}