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;
typedef int (*uniopi)(int);
typedef long long (*uniopl)(long long);
typedef int (*binopi)(int, int);
typedef long long (*binopl)(long long, long long);
typedef int (*tupopi)(int, int, int);
typedef long long (*tupopl)(long long, long long, long long);
const long double PI = acos(0) * 2;
const int INF = 0x3f2f1f0f;
const long long LINF = 1ll * INF * INF;
int X4[] = {-1, 0, 1, 0}, Y4[] = {0, -1, 0, 1};
int X8[] = {-1, -1, -1, 0, 1, 1, 1, 0}, Y8[] = {-1, 0, 1, 1, 1, 0, -1, -1};
template <typename T>
inline bool uax(T& a, T b) {
  return a < b ? (a = b, 1) : 0;
}
template <typename T>
inline bool uin(T& a, T b) {
  return a > b ? (a = b, 1) : 0;
}
template <typename T1, typename T2>
istream& operator>>(istream& is, pair<T1, T2>& p) {
  return is >> p.first >> p.second;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& p) {
  return os << '(' << p.first << ", " << p.second << ")";
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  for (int i = 0; i < (int)v.size(); i++) {
    os << v[i];
    if (i + 1 < (int)v.size()) os << ' ';
  }
  return os;
}
void filll(int xs, int xf, int ys, int yf, int u, vector<vector<int> >& a) {
  int c = 0;
  for (int i = xs; i < xf; i++)
    for (int j = ys; j < yf; j++) a[i][j] = 4 * (c++) + u;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout << std::setprecision(10);
  cout << fixed;
  int n;
  cin >> n;
  int m = n / 2;
  vector<vector<int> > a(n, vector<int>(n, 0));
  filll(0, m, 0, m, 0, a);
  filll(0, m, m, n, 1, a);
  filll(m, n, 0, m, 2, a);
  filll(m, n, m, n, 3, a);
  for (auto& v : a) {
    cout << v << endl;
  }
  return 0;
}