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 a[1000][1000];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long int n;
  cin >> n;
  a[0][0] = 8;
  a[0][1] = 9;
  a[0][2] = 1;
  a[0][3] = 13;
  a[1][0] = 3;
  a[1][1] = 12;
  a[1][2] = 7;
  a[1][3] = 5;
  a[2][0] = 0;
  a[2][1] = 2;
  a[2][2] = 4;
  a[2][3] = 11;
  a[3][0] = 6;
  a[3][1] = 10;
  a[3][2] = 15;
  a[3][3] = 14;
  int add = 16;
  for (int i = 0; i < n; i += 4) {
    for (int j = 0; j < n; j += 4) {
      if (i == 0 && j == 0) {
        continue;
      }
      a[i][j] = 8 + add;
      a[i][j + 1] = 9 + add;
      a[i][2 + j] = 1 + add;
      a[i][3 + j] = 13 + add;
      a[i + 1][j] = 3 + add;
      a[i + 1][1 + j] = 12 + add;
      a[i + 1][2 + j] = 7 + add;
      a[i + 1][3 + j] = 5 + add;
      a[i + 2][j] = 0 + add;
      a[i + 2][1 + j] = 2 + add;
      a[i + 2][2 + j] = 4 + add;
      a[i + 2][3 + j] = 11 + add;
      a[i + 3][j] = 6 + add;
      a[i + 3][1 + j] = 10 + add;
      a[i + 3][2 + j] = 15 + add;
      a[i + 3][3 + j] = 14 + add;
      add += 16;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) cout << a[i][j] << ' ';
    cout << '\n';
  }
  return 0;
}