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 mx = 1100;
int m[mx][mx], n;
void dfs(int x, int y, int n, int k) {
  if (n == 4) {
    int num = n * n - 1;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) {
        m[x + i - 1][y + j - 1] = num + k;
        num--;
      }
    return;
  }
  int i = 0;
  for (i = 0;; i++) {
    if (x + 4 * i > n) break;
    for (int j = 0;; j++) {
      if (y + 4 * j > n)
        break;
      else
        dfs(x + 4 * i, y + 4 * j, 4, k), k += 16;
    }
  }
}
int main() {
  cin >> n;
  dfs(1, 1, n, 0);
  for (int i = n; i >= 1; i--) {
    for (int j = n; j >= 1; j--) cout << m[i][j] << " ";
    cout << endl;
  }
  return 0;
}
