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 N = 200005;
const long long M = 1000000007;
long long pow1(long long a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % M;
    }
    a = (a * a) % M;
    b >>= 1;
  }
  return res % M;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long T;
  T = 1;
  while (T--) {
    long long n;
    cin >> n;
    long long ans[n][n];
    long long num = 0;
    for (long long i = 0; i < n; i++) {
      if (i % 2 == 1) {
        num = ans[0][i - 1] + 1;
      } else {
        num = n * (i);
      }
      long long num1 = num;
      for (long long j = 0; j < n; j++) {
        ans[j][i] = num1;
        num1 += 2;
      }
    }
    set<long long> se;
    for (long long i = 0; i < n; i++) {
      long long num = 0;
      for (long long j = 0; j < n; j++) num = num ^ ans[i][j];
      se.insert(num);
    }
    for (long long i = 0; i < n; i++) {
      long long num = 0;
      for (long long j = 0; j < n; j++) num = num ^ ans[j][i];
      se.insert(num);
    }
    for (long long i = 0; i < n; i++) {
      for (long long j = 0; j < n; j++) cout << ans[i][j] << " ";
      cout << endl;
    }
  }
}
