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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int small_grid[4][4] = {8, 9, 1, 13, 3, 12, 7, 5, 0, 2, 4, 11, 6, 10, 15, 14};
int grid[1010][1010];
int main() {
  int n;
  while (scanf("%d", &n) > 0) {
    for (int i = 0, k = 0; i < n; i += 4) {
      for (int j = 0; j < n; j += 4, k += 16) {
        for (int ii = 0; ii < 4; ii++) {
          for (int jj = 0; jj < 4; jj++) {
            grid[i + ii][j + jj] = small_grid[ii][jj] + k;
          }
        }
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        printf("%d%c", grid[i][j], j + 1 == n ? '\n' : ' ');
      }
    }
    continue;
    for (int i = 0; i < n; i++) {
      int row_xor = 0;
      for (int j = 0; j < n; j++) {
        row_xor ^= grid[i][j];
      }
      printf("%d\n", row_xor);
    }
    for (int j = 0; j < n; j++) {
      int col_xor = 0;
      for (int i = 0; i < n; i++) {
        col_xor ^= grid[i][j];
      }
      printf("%d\n", col_xor);
    }
  }
}