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 = 1005;
int ans[N][N];
vector<int> sums[N * N];
int main() {
  ios ::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n * n; i += 4) {
    for (int j = 0; j < 4; j++) {
      sums[i / 4].push_back(i + j);
    }
  }
  memset(ans, -1, sizeof ans);
  int block = 0;
  int indices[4][4] = {{0, 1, 2, 3}, {2, 3, 0, 1}, {1, 2, 3, 0}, {3, 0, 1, 2}};
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (ans[i][j] == -1) {
        int indices_block = -1;
        if (i % 4 == 0 && j % 4 == 0) {
          indices_block = 0;
        } else if (i % 4 == 1 && j % 4 == 0) {
          indices_block = 1;
        } else if (i % 4 == 0 && j % 4 == 1) {
          indices_block = 2;
        } else {
          indices_block = 3;
        }
        ans[i][j] = sums[block][indices[indices_block][0]];
        ans[i + 2][j] = sums[block][indices[indices_block][1]];
        ans[i][j + 2] = sums[block][indices[indices_block][2]];
        ans[i + 2][j + 2] = sums[block][indices[indices_block][3]];
        block++;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << ans[i][j] << ' ';
    }
    cout << '\n';
  }
}