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.util.*;
public class contest11 {
	static int val=0;
	static Scanner scn = new Scanner(System.in);
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int n = scn.nextInt();
		int[][] grid = new int[n][n];
		
		for( int row=0;row<n;row+=4){
			for( int col=0;col<n;col+=4)
			fill4by4box(grid,row,col);
		}
		
		for(int i=0;i<n;i++) {
			StringBuilder sb = new StringBuilder();
			for(int j=0;j<n;j++) {
				sb.append(grid[i][j]+" ");
			}
			System.out.println(sb);
		}
	}
	
	public static void fill4by4box(int[][] grid, int row, int col) {
		for( int i=row;i<row+4;i++) {
			for(int j=col;j<col+4;j++) {
				//System.out.println(i+" "+j+" "+val);
				grid[i][j]=val++;
			}
		}
	}

}
 	 	 	     	 	     	 	 			 		