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 double PI = 2 * acos(0.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
mt19937_64 rng_64(chrono::steady_clock::now().time_since_epoch().count());
const string DIGITS = "0123456789";
const string ALPH = "abcdefghijklmnopqrstuvwxyz";
template <class T0, class T1>
inline ostream &operator<<(ostream &out, pair<T0, T1> &a) {
  return out << "{" << a.first << ", " << a.second << "}";
}
template <class T0, class T1, class T2>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << "}";
}
template <class T0, class T1, class T2, class T3>
inline ostream &operator<<(ostream &out, tuple<T0, T1, T2, T3> &a) {
  return out << "{" << get<0>(a) << ", " << get<1>(a) << ", " << get<2>(a)
             << ", " << get<3>(a) << "}";
}
template <class T>
inline ostream &operator<<(ostream &out, vector<T> &a) {
  out << "[";
  for (int i = 0; i < (int)(a.size()); ++i)
    out << a[i] << vector<string>{", ", "]  "}[i + 1 == a.size()];
  return out;
}
void smain();
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(12) << fixed;
  smain();
  return 0;
}
const int S = 1010;
const int N = S;
int a[N][N];
void smain() {
  int n;
  cin >> n;
  bool ok = true;
  for (int i = 0; i < (int)(n); ++i)
    for (int j = 0; j < (int)(n); ++j) {
      a[i][j] = (i % 4) * 4 + (j % 4);
      a[i][j] |= ((i / 4) * (n / 4) + (j / 4)) << 4;
    }
  for (int i = 0; i < n; ++i) {
    int xr = 0;
    for (int j = 0; j < n; ++j) {
      xr ^= a[i][j];
    }
    if (xr) ok = false;
  }
  for (int j = 0; j < n; ++j) {
    int xr = 0;
    for (int i = 0; i < n; ++i) {
      xr ^= a[i][j];
    }
    if (xr) ok = false;
  }
  set<int> used;
  for (int i = 0; i < (int)(n); ++i)
    for (int j = 0; j < (int)(n); ++j) {
      used.insert(a[i][j]);
    }
  ok &= used.size() == n * n;
  if (false) cerr << ("ok") << " = " << (ok) << endl;
  ;
  if (ok || true) {
    for (int i = 0; i < (int)(n); ++i) {
      for (int j = 0; j < (int)(n); ++j) {
        cout << a[i][j] << ' ';
      }
      cout << '\n';
    }
    return;
  }
}