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 n;
int a[N][N] = {
    {8, 9, 1, 13},
    {3, 12, 7, 5},
    {0, 2, 4, 11},
    {6, 10, 15, 14},
};
bool check() {
  int xxor = -1;
  for (int i = 0; i < n; ++i) {
    int xxx = 0;
    for (int j = 0; j < n; ++j) {
      xxx ^= a[i][j];
    }
    if (xxor == -1 || xxor == xxx)
      xxor = xxx;
    else
      return false;
  };
  xxor = -1;
  for (int j = 0; j < n; ++j) {
    int xxx = 0;
    for (int i = 0; i < n; ++i) {
      xxx ^= a[i][j];
    }
    if (xxor == -1 || xxor == xxx)
      xxor = xxx;
    else
      return false;
  };
  return true;
}
int main() {
  scanf("%d", &n);
  int m = n / 4;
  for (int i = 0; i < m; ++i) {
    for (int j = 0; j < m; ++j) {
      for (int k = 4 * i; k < 4 * (i + 1); ++k) {
        for (int p = 4 * j; p < 4 * (j + 1); ++p) {
          a[k][p] = a[k - 4 * i][p - 4 * j] + 16 * (i * m + j);
        }
      }
    }
  }
  assert(check());
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cout << a[i][j] << ' ';
    }
    cout << endl;
  }
  return 0;
}