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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("unswitch-loops")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("rename-registers")
#pragma GCC optimize("move-loop-invariants")
#pragma GCC optimize("function-sections")
#pragma GCC optimize("data-sections")
#pragma GCC optimize("branch-target-load-optimize")
#pragma GCC optimize("branch-target-load-optimize2")
#pragma GCC optimize("btr-bb-exclusive")
#pragma GCC target("sse2")
#pragma GCC target("sse3")
#pragma GCC target("sse4.1")
#pragma GCC target("sse4.2")
#pragma GCC target("popcnt")
#pragma GCC target("abm")
#pragma GCC target("mmx")
#pragma GCC target("tune=native")
using namespace std;
void addcopy(vector<vector<int>>& t, vector<vector<int>>& r, int x, int y,
             int sum) {
  for (int i = 0; i < 4; ++i) {
    for (int j = 0; j < 4; ++j) {
      r[x + i][j + y] = t[i][j] + sum;
    }
  }
}
int main() {
  cin.tie();
  cout.tie();
  ios_base::sync_with_stdio(0);
  vector<vector<int>> t = {
      {8, 9, 1, 13}, {3, 12, 7, 5}, {0, 2, 4, 11}, {6, 10, 15, 14}};
  int n;
  cin >> n;
  vector<vector<int>> r(n, vector<int>(n));
  int k = n / 4;
  for (int i = 0; i < k; ++i) {
    for (int j = 0; j < k; ++j) {
      addcopy(t, r, i * 4, j * 4, 16 * (j + i * k));
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) cout << r[i][j] << ' ';
    cout << '\n';
  }
}