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;
const int mod = 1e9 + 7;
const int N = 3e5 + 5;
int power(int a, int b) {
  if (!b) return 1;
  int c = power(a, b / 2);
  c = (1LL * c * c) % mod;
  if (b % 2) c = (1LL * c * a) % mod;
  return c;
}
void done() {}
int n;
bool valid(int i, int j) { return (i >= 0 && i < n && j >= 0 && j < n); }
void solve() {
  int n;
  cin >> n;
  int c[n + 1][n + 1];
  int k = 0;
  for (int i = 1; i <= n / 2; ++i) {
    for (int j = 1; j <= n / 2; ++j) {
      c[i][j] = 4 * k + 1;
      c[i][j + n / 2] = 4 * k + 2;
      c[i + n / 2][j] = 4 * k + 3;
      c[i + n / 2][j + n / 2] = 4 * k;
      ++k;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      cout << c[i + 1][j + 1] << ' ';
    }
    cout << '\n';
  }
}
void another() {
  int n;
  cin >> n;
  int s[n][n];
  for (int i = 0; i < n; ++i) {
    for (int ad = 0; ad < n; ++ad) {
      cin >> s[i][ad];
    }
  }
  int a = s[1][0] - '0';
  int b = s[0][1] - '0';
  int c = s[n - 1][n - 2] - '0';
  int d = s[n - 2][n - 1] - '0';
  vector<pair<int, int> > ans;
  if (a == b) {
  }
}
void test_case() {
  int t;
  cin >> t;
  while (t--) another();
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  solve();
}