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;
const int MXN = 2e5 + 30;
const int MAXN = 1e3 + 560;
const long long INF = 1e9 + 7;
const long long BINF = 1e16;
const int mod = 1e9 + 7;
const long double EPS = 1e-15;
const long long dx[] = {0, 0, 1, -1};
const long long dy[] = {1, -1, 0, 0};
int n;
int a[MAXN][MAXN];
int used[MXN * 10];
int m;
vector<int> v;
int sz;
int main() {
  cin >> n;
  a[0][0] = 8, a[0][1] = 9, a[0][2] = 1, a[0][3] = 13;
  a[1][0] = 3, a[1][1] = 12, a[1][2] = 7, a[1][3] = 5;
  a[2][0] = 0, a[2][1] = 2, a[2][2] = 4, a[2][3] = 11;
  a[3][0] = 6, a[3][1] = 10, a[3][2] = 15, a[3][3] = 14;
  for (int mask = 1; mask < n * n; mask++) {
    if (mask * 16 + 15 < n * n) v.push_back(mask);
  }
  for (int i = 0; i < n / 4; i++) {
    for (int j = 0; j < n / 4; j++) {
      if (i + j == 0) continue;
      for (int x = 0; x < 4; x++) {
        for (int y = 0; y < 4; y++) {
          a[i * 4 + x][j * 4 + y] = v[sz] * 16 + a[x][y];
        }
      }
      sz++;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << a[i][j] << ' ';
    }
    cout << '\n';
  }
  return 0;
}