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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Kraken
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    Scanner in = new Scanner(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    TaskC solver = new TaskC();
    solver.solve(1, in, out);
    out.close();
  }

  static class TaskC {

    public void solve(int testNumber, Scanner in, PrintWriter out) {
      int n = in.nextInt();
      long curr = 0;
      long[][] mat = new long[n][n];
      for (int i = 0; i < n / 2; i++) {
        for (int j = 0; j < n / 2; j++) {
          mat[i][j] = 4 * curr + 1;
          mat[i][j + n / 2] = 4 * curr + 2;
          mat[i + n / 2][j] = 4 * curr + 3;
          mat[i + n / 2][j + n / 2] = 4 * curr;

          curr++;
        }
      }

      for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
          out.print(mat[i][j] + " ");
        }
        out.println();
      }
    }

  }
}