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;
template <typename T>
inline void read(T &t) {
  int f = 0, c = getchar();
  t = 0;
  while (!isdigit(c)) f |= c == '-', c = getchar();
  while (isdigit(c)) t = t * 10 + c - 48, c = getchar();
  if (f) t = -t;
}
template <typename T, typename... Args>
inline void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
int ans[1005][1005];
int main() {
  int n;
  read(n);
  int sum = 0;
  for (int i = 1; i <= n / 4; i++) {
    for (int j = 1; j <= n / 4; j++) {
      ans[i * 4][j * 4] = 8 + sum;
      ans[i * 4][j * 4 + 1] = 9 + sum;
      ans[i * 4][j * 4 + 2] = 1 + sum;
      ans[i * 4][j * 4 + 3] = 13 + sum;
      ans[i * 4 + 1][j * 4] = 3 + sum;
      ans[i * 4 + 1][j * 4 + 1] = 12 + sum;
      ans[i * 4 + 1][j * 4 + 2] = 7 + sum;
      ans[i * 4 + 1][j * 4 + 3] = 5 + sum;
      ans[i * 4 + 2][j * 4] = 0 + sum;
      ans[i * 4 + 2][j * 4 + 1] = 2 + sum;
      ans[i * 4 + 2][j * 4 + 2] = 4 + sum;
      ans[i * 4 + 2][j * 4 + 3] = 11 + sum;
      ans[i * 4 + 3][j * 4] = 6 + sum;
      ans[i * 4 + 3][j * 4 + 1] = 10 + sum;
      ans[i * 4 + 3][j * 4 + 2] = 15 + sum;
      ans[i * 4 + 3][j * 4 + 3] = 14 + sum;
      sum += 16;
    }
  }
  for (int i = 4; i <= n + 3; i++) {
    for (int j = 4; j <= n + 3; j++) {
      cout << ans[i][j] << " ";
    }
    cout << endl;
  }
}
