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;
void solve() {
  int n;
  cin >> n;
  vector<vector<int> > ans(n, vector<int>(n));
  int start = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = start, idx = 0; idx < n; j += 4, idx++) ans[i][idx] = j;
    if ((i + 1) % 4 == 0) {
      start = n * (i + 1);
    } else {
      start++;
    }
  }
  set<int> x0rs;
  for (int i = 0; i < n; ++i) {
    int x0r = 0;
    for (int j = 0; j < n; ++j) {
      x0r ^= ans[i][j];
    }
    x0rs.insert(x0r);
    x0r = 0;
    for (int j = 0; j < n; ++j) {
      x0r ^= ans[j][i];
    }
    x0rs.insert(x0r);
  }
  cerr << "x0rs:" << endl;
  for (int el : x0rs) {
    cerr << el << " ";
  }
  cerr << endl;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cout << ans[i][j] << " ";
    }
    cout << endl;
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
  return 0;
}
