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:
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF1208C extends PrintWriter {
	CF1208C() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1208C o = new CF1208C(); o.main(); o.flush();
	}

	void main() {
		int n = sc.nextInt();
		int[][] aa = new int[n][n];
		for (int i4 = 0; i4 < n / 4; i4++)
			for (int j4 = 0; j4 < n / 4; j4++) {
				int k4 = i4 * (n / 4) + j4;
				for (int i = 0; i < 4; i++)
					for (int j = 0; j < 4; j++) {
						int k = i * 4 + j;
						aa[i4 * 4 + i][j4 * 4 + j] = k4 * 16 + k;
					}
			}
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++)
				print(aa[i][j] + " ");
			println();
		}
	}
}