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.
#include <bits/stdc++.h>
using namespace std;
const int INF = ~(1 << 31);
const int NINF = 1 << 31;
const int N = 1e3 + 5;
int a[N][N];
int b[4][4] = {8, 9, 1, 13, 3, 12, 7, 5, 0, 2, 4, 11, 6, 10, 15, 14};
int n, m;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i <= 3; ++i)
    for (int j = 0; j <= 3; ++j) a[i][j] = b[i][j];
  int k = n / 4, c = -1;
  for (int i = 0; i <= k - 1; ++i) {
    for (int j = 0; j <= k - 1; ++j) {
      c++;
      for (int z1 = 0; z1 <= 3; ++z1)
        for (int z2 = 0; z2 <= 3; ++z2)
          a[i * 4 + z1][j * 4 + z2] = b[z1][z2] + 4 * 4 * c;
    }
  }
  for (int i = 0; i <= n - 1; ++i) {
    for (int j = 0; j <= n - 1; ++j) {
      cout << a[i][j];
      if (j == n - 1)
        cout << endl;
      else
        cout << ' ';
    }
  }
  return 0;
}
