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>
long long mod(long long x) {
  if (x < 0) return -x;
  return x;
}
long long expo(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y % 2 == 1) res = (res * x);
    y = y >> 1;
    x = x * x;
  }
  return res;
}
using namespace std;
const int MAXN = 500005;
priority_queue<int, vector<int>, greater<int> > pq;
vector<long long> v;
long long a[1003][1003], b[1000000];
int main() {
  ios_base::sync_with_stdio(false);
  long long n = 0, t = 0, m = 0, k = 0, x = 0, y = 0;
  cin >> n;
  for (long long i = 0; i < n / 4; i++) {
    for (long long o = 0; o < n / 4; o++) {
      for (long long j = 0; j < 4; j++) {
        for (long long k = 0; k < 4; k++) {
          a[4 * i + j][4 * o + k] = m;
          m++;
        }
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      cout << a[i][j] << " ";
    }
    cout << endl;
  }
  return 0;
}