Problem: Let us define a magic grid to be a square matrix of integers of size n × n, satisfying the following conditions. 

  * All integers from 0 to (n^2 - 1) inclusive appear in the matrix exactly once. 
  * [Bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of all elements in a row or a column must be the same for each row and column. 



You are given an integer n which is a multiple of 4. Construct a magic grid of size n × n.

Input

The only line of input contains an integer n (4 ≤ n ≤ 1000). It is guaranteed that n is a multiple of 4.

Output

Print a magic grid, i.e. n lines, the i-th of which contains n space-separated integers, representing the i-th row of the grid.

If there are multiple answers, print any. We can show that an answer always exists.

Examples

Input


4


Output


8 9 1 13
3 12 7 5
0 2 4 11
6 10 15 14

Input


8


Output


19 55 11 39 32 36 4 52
51 7 35 31 12 48 28 20
43 23 59 15 0 8 16 44
3 47 27 63 24 40 60 56
34 38 6 54 17 53 9 37
14 50 30 22 49 5 33 29
2 10 18 46 41 21 57 13
26 42 62 58 1 45 25 61

Note

In the first example, XOR of each row and each column is 13.

In the second example, XOR of each row and each column is 60.

Solution (in JAVA):

Solution:
import java.util.Arrays;
import java.util.Scanner;

public class Test {




    public static void main(String args[] ) throws Exception {

        Scanner s = new Scanner(System.in);

        int  n = s.nextInt();

        long a[][] = new long[n][n];


        int c = 0;
        for(int i=0;i<n/2;i++){
            for(int j=0;j<n/2;j++){

                a[i][j] = 4*c;
                a[i][j+n/2] = 4*c + 1;
                a[i+n/2][j] = 4*c + 2;
                a[i+n/2][j+n/2] = 4*c + 3;


                c++;
            }

        }

        StringBuilder sb = new StringBuilder();
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                sb.append(a[i][j]+" ");
            }
            sb.append("\n");
        }
       System.out.println(sb);

    }

}