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;
using ll = long long;
vector<vector<ll>> solve(ll n) {
  ll next = 0;
  vector<vector<ll>> G(n, vector<ll>(n, 0));
  for (ll r = 0; r < n / 2; r++) {
    for (ll c = 0; c < n / 2; c++) {
      G[r][c] = next * 4;
      G[r][c + n / 2] = next * 4 + 1;
      G[r + n / 2][c] = next * 4 + 2;
      G[r + n / 2][c + n / 2] = next * 4 + 3;
      next++;
    }
  }
  for (ll r = 0; r < n; r++) {
    ll sum = 0;
    for (ll c = 0; c < n; c++) {
      sum ^= G[r][c];
    }
    assert(sum == 0);
  }
  for (ll c = 0; c < n; c++) {
    ll sum = 0;
    for (ll r = 0; r < n; r++) {
      sum ^= G[r][c];
    }
    assert(sum == 0);
  }
  return G;
}
int main() {
  ll n;
  cin >> n;
  assert(n % 4 == 0);
  vector<vector<ll>> G = solve(n);
  for (ll r = 0; r < n; r++) {
    for (ll c = 0; c < n; c++) {
      cout << G[r][c];
      if (c == n - 1) {
        cout << endl;
      } else {
        cout << " ";
      }
    }
  }
}