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:
//package CF1208;

import java.io.*;
import java.util.*;

public class C {

	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);
		PrintWriter cout = new PrintWriter(System.out);
		
		int n = cin.nextInt();
		int[][] a = new int[1003][1003];
		int cnt = 0;
		
		for (int i = 0; i <= n/4-1; ++i) {
			for (int j = 0; j <= n/4-1; ++j) {
				for (int ii = i*4+1; ii <= (i+1)*4; ++ii) {
					for (int jj = j*4+1; jj <= (j+1)*4; ++jj)
						a[ii][jj] = cnt++;
				}
			}
		}
		for (int i = 1; i <= n; ++i) {
			for (int j = 1; j <= n; ++j) {
				cout.print(a[i][j]);
				cout.print(" ");
			}
			cout.print("\n");
		}
		cout.close();
	}

}