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;
void FREOPEN() {}
inline double Time() { return (clock() * 1.0) / CLOCKS_PER_SEC; }
const int N = (int)2000, inf = (int)2e9, MOD = (int)1e9 + 7, alph = 26;
char CH[N];
const long long INF = (long long)1e18;
const int dx[] = {1, -1, 0, 0, -1, 1, -1, 1};
const int dy[] = {0, 0, 1, -1, -1, 1, 1, -1};
const int dp[] = {8, 9, 1, 13, 3, 12, 7, 5, 0, 2, 4, 11, 6, 10, 15, 14};
int a[N][N];
int main() {
  FREOPEN();
  int n;
  cin >> n;
  int num = 0;
  for (int bi = 0; bi < n; bi += 4) {
    for (int bj = 0; bj < n; bj += 4) {
      for (int i = bi; i < bi + 4; i++) {
        for (int j = bj; j < bj + 4; j++) {
          int p = (i - bi) * 4 + (j - bj);
          a[i][j] = (dp[p] ^ num);
        }
      }
      num += 16;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) printf("%d ", a[i][j]);
    printf("\n");
  }
  return 0;
}