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;
const long long MOD = 1000000007;
int fx[] = {0, 0, -1, 1, +1, -1, -1, +1};
int fy[] = {1, -1, 0, 0, +1, +1, -1, -1};
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return a.first > b.first;
}
template <typename T>
inline T Bigmod(T base, T power, T MOD) {
  T ret = 1;
  while (power) {
    if (power & 1) ret = (ret * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  int n;
  while (cin >> n) {
    int lim = n / 2;
    int arr[lim][lim];
    int trace = 0;
    for (int i = 0; i < lim; i++) {
      for (int j = 0; j < lim; j++) {
        arr[i][j] = trace;
        trace++;
      }
    }
    for (int i = 0; i < lim; i++) {
      for (int j = 0; j < n; j++) {
        if (j < lim) {
          cout << arr[i][j] * 4 << " ";
        } else
          cout << ((arr[i][j - lim] * 4) + 1) << " ";
      }
      cout << endl;
    }
    for (int i = lim; i < n; i++) {
      for (int j = 0; j < n; j++) {
        if (j < lim)
          cout << ((arr[i - lim][j] * 4) + 2) << " ";
        else
          cout << ((arr[i - lim][j - lim] * 4) + 3) << " ";
      }
      cout << endl;
    }
  }
  return 0;
}
