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 inf = 0x3f3f3f3f, mod = 1000000007;
const long long INF = 0x3f3f3f3f3f3f3f3fLL;
const double pi = acosl(-1.), eps = 1e-9;
inline int power(int a, int b, int m = mod, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int ans[1111][1111];
int main() {
  int n;
  cin >> n;
  n /= 2;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      int t = (i * n + j) * 4;
      if ((i & 1) == (j & 1)) {
        ans[i * 2][j * 2] = t + 0;
        ans[i * 2][j * 2 + 1] = t + 1;
        ans[i * 2 + 1][j * 2] = t + 2;
        ans[i * 2 + 1][j * 2 + 1] = t + 3;
      } else {
        ans[i * 2][j * 2] = t + 0;
        ans[i * 2][j * 2 + 1] = t + 2;
        ans[i * 2 + 1][j * 2] = t + 1;
        ans[i * 2 + 1][j * 2 + 1] = t + 3;
      }
    }
  }
  n *= 2;
  for (int i = 0; i < n; i++, puts("")) {
    for (int j = 0; j < n; j++) printf("%d ", ans[i][j]);
  }
  return 0;
}