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;
const int N = 1111;
int a[N][N];
int b[N][N] = {{0, 2, 8, 10}, {1, 3, 9, 11}, {4, 6, 12, 14}, {5, 7, 13, 15}};
int n;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int pi = (0); pi <= (n / 4 - 1); pi++) {
    for (int pj = (0); pj <= (n / 4 - 1); pj++) {
      for (int ii = (0); ii <= (3); ii++) {
        for (int jj = (0); jj <= (3); jj++) {
          int i = pi * 4 + ii;
          int j = pj * 4 + jj;
          a[i][j] = 16 * (pi * (n / 4) + pj) + b[ii][jj];
        }
      }
    }
  }
  for (int i = (0); i <= (n - 1); i++) {
    for (int j = (0); j <= (n - 1); j++) {
      cout << a[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}