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;
long long int inf = 1e9 + 7;
bool check(map<long long int, long long int> ma) {
  for (auto i = ma.begin(); i != ma.end(); ++i) {
    if (i->second < 2) continue;
    return false;
  }
  return true;
}
bool check1(long long int mid, vector<long long int> v,
            map<long long int, long long int> ma) {
  if (mid == 0) {
    if (check(ma)) return true;
    return false;
  }
  for (long long int j = 0; j < mid; ++j) ma[v[j]]--;
  if (check(ma)) return true;
  for (long long int j = (mid); j <= (v.size() - mid + 1); ++j) {
    ma[v[j]]--;
    ma[v[j - mid]]++;
    if (check(ma)) return true;
  }
  return false;
}
vector<vector<long long int>> ff(long long int k) {
  vector<vector<long long int>> v(4, vector<long long int>(4));
  for (long long int i = 0; i < 4; ++i) {
    for (long long int j = 0; j < 4; ++j) v[i][j] = k++;
  }
  return v;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int k = 0;
  long long int ans[n][n];
  for (long long int i = 0; i < n / 4; ++i) {
    for (long long int j = 0; j < n / 4; ++j) {
      vector<vector<long long int>> v = ff(k);
      for (long long int i1 = 0; i1 < 4; ++i1) {
        for (long long int j1 = 0; j1 < 4; ++j1)
          ans[i1 + i * 4][j1 + j * 4] = v[i1][j1];
      }
      k += 16;
    }
  }
  for (long long int i = 0; i < n; ++i) {
    for (long long int j = 0; j < n; ++j) cout << ans[i][j] << " ";
    cout << "\n";
  }
  return 0;
}
