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.*;
import java.io.*;
import java.lang.*;
import java.math.*;
public class C {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int n = Integer.parseInt(bf.readLine());
        int[][] a = new int[n][n];
        for(int i=0; i<n; i++) {
          for(int j=0; j<n; j++) {
            a[i][j] = ((i%(n/4))*n/4 + (j%(n/4)))*16 + (i/(n/4)*4+(j/(n/4)));
          }
        }
        for(int i=0; i<n; i++) {
          StringBuilder sb = new StringBuilder();
          for(int j=0; j<n; j++) {
            sb.append(a[i][j] + " ");

          }
          out.println(sb.toString());
        }
        // StringTokenizer st = new StringTokenizer(bf.readLine());
        // int[] a = new int[n]; for(int i=0; i<n; i++) a[i] = Integer.parseInt(st.nextToken());
        // int n = Integer.parseInt(st.nextToken());

        out.close(); System.exit(0);
    }
}