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.
#include <bits/stdc++.h>
using namespace std;
const int MAX_N = 1e6 + 5;
const double delta = 1e-8;
const long long mod = 998244853;
const int magic_const = 1e9;
mt19937 gen(54323244234);
long long get_rand(long long x) { return (gen() % x + x) % x; }
bool get_last_twin(vector<int> &a) {
  unordered_set<int> s;
  int answer = -1;
  for (int i = 0; i < a.size(); i++) {
    if (s.find(a[i]) != s.end()) {
      return false;
    } else {
      s.insert(a[i]);
    }
  }
  return true;
}
vector<int> get_vector_a(vector<int> &a, int l, int r) {
  vector<int> answer;
  for (int i = 0; i < l; i++) {
    answer.push_back(a[i]);
  }
  for (int i = r + 1; i < a.size(); i++) {
    answer.push_back(a[i]);
  }
  return answer;
}
void solve() {
  int n;
  cin >> n;
  int max_n = n * n;
  vector<int> xor_d(n);
  vector<int> xor_t(n);
  vector<vector<int>> a(n, vector<int>(n));
  int path_size = n * 4;
  for (int i = 0; i < max_n; i++) {
    int path = i / path_size;
    int new_x = path * 4 + i % 4;
    int new_y = (i - path * path_size) / 4;
    a[new_x][new_y] = i;
    xor_d[new_x] ^= i;
    xor_t[new_y] ^= i;
  }
  for (auto &i : a) {
    for (auto j : i) cout << j << " ";
    cout << "\n";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand(58164);
  solve();
}
