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 boolean cross(int masB[][],int i, int j,int m,int n){
        for(int k=0;k<n;k++) if (masB[i][k]==0) return false;
        for(int k=0;k<m;k++) if (masB[k][j]==0) return false;
        return true;
    }

    static void cross2(int masB[][],int i, int j,int m,int n){
        for(int k=0;k<n;k++) masB[i][k]=1;
        for(int k=0;k<m;k++) masB[k][j]=1;
    }

    public static void main(String[] args) {
	    Scanner S=new Scanner(System.in);
        int m,n;
        m=S.nextInt();
        n=S.nextInt();
        int masA[][]=new int[m][n];
        int masB[][]=new int[m][n];
        int masC[][]=new int[m][n];
        for(int i=0;i<m;i++) for(int j=0;j<n;j++) masB[i][j]=S.nextInt();
        for(int i=0;i<m;i++) for(int j=0;j<n;j++) if (cross(masB,i,j,m,n)) masA[i][j]=1;
        else masA[i][j]=0;
        for(int i=0;i<m;i++) for(int j=0;j<n;j++) masC[i][j]=0;
        for(int i=0;i<m;i++) for(int j=0;j<n;j++) if(masA[i][j]==1) cross2(masC,i,j,m,n);
        for(int i=0;i<m;i++) for(int j=0;j<n;j++) if (masC[i][j]!=masB[i][j]) {
            System.out.print("NO");
            return;
        }
        System.out.println("YES");
        for(int i=0;i<m;i++) {
            for (int j = 0; j < n-1; j++) System.out.print(masA[i][j] + " ");
            System.out.println(masA[i][n-1]);
        }
    }
}