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:
/**
 * @author Finn Lidbetter
 */
import java.util.*;
import java.io.*;
import java.awt.geom.*;

public class TaskC {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringBuilder sb = new StringBuilder();
    
    int[][] base = new int[][]{{8,9,1,13},{3,12,7,5},{0,2,4,11}, {6,10,15,14}};

    int n = Integer.parseInt(br.readLine());
    int[][] arr = new int[n][n];
    int prefix = 0;
    for (int i=0; i<n/4; i++) {
      for (int j=0; j<n/4; j++) {
        for (int k=0; k<4; k++) {
          for (int l=0; l<4; l++) {
            arr[4*i+k][4*j+l] = base[k][l] + (prefix<<4);
          }
        }
        prefix++;
      }
    }
    /*
    int xor = -1;
    for (int i=0; i<n; i++) {
      int row = 0;
      for (int j=0; j<n; j++) {
        row ^= arr[j][i];
      }
      if (xor==-1) {
        xor = row;
      }
      System.out.println(row);
      if (row!=xor) {
        System.out.println("ERROR");
      }
    }
    for (int i=0; i<n; i++) {
      int col = 0;
      for (int j=0; j<n; j++) {
        col ^= arr[j][i];
      }
      System.out.println(col);
      if (col!=xor) {
        System.out.println("ERROR");
      }
    }
    */
    for (int i=0; i<n; i++) {
      for (int j=0; j<n; j++) {
        sb.append(arr[i][j]);
        sb.append((j==n-1) ? "\n" : " ");
      }
    }
    System.out.print(sb);

  }
}