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.
import java.io.*;
import java.util.*;
public class Solution {
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	//debug
	static void debug(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
	//Main
	public static void main(String args[]) {
		int test=1;
		//test=sc.nextInt();
		while(test-->0) {
			//Think before code, it won't hurt
			int n=sc.nextInt(),a[][]=new int[n][n];
			int cur=0;
			for(int i=0;i<n/2;i++) {
				for(int j=0;j<n/2;j++) {
					a[i][j]=4*cur+1;
					a[i][j+n/2]=4*cur+2;
					a[i+n/2][j]=4*cur+3;
					a[i+n/2][j+n/2]=4*cur;
					cur++;
				}
			}
			for(int i=0;i<n;i++) {
				for(int j=0;j<n;j++) out.print(a[i][j]+" ");
				out.println();
			}
		}
		out.flush();
		out.close();
	}
}
