#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
template <class T>
using V = vector<T>;
const ll MOD = 1e9 + 7;
const int MX = 2e+5;
const ll INF = 1e18;
const db PI = acos((db)-1);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
constexpr int p2(int x) { return 1 << x; }
constexpr int msk2(int x) { return p2(x) - 1; }
void setIO() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
}
int main() {
  setIO();
  int n;
  cin >> n;
  V<vi> res(n, vi(n));
  for (int i = (0); i < (n / 2); ++i) {
    for (int j = (0); j < (n / 2); ++j) {
      res[i][j] = 2 * n * i + 4 * j;
      res[i][j + n / 2] = res[i][j] + 1;
      res[i + n / 2][j] = res[i][j] + 2;
      res[i + n / 2][j + n / 2] = res[i][j] + 3;
    }
  }
  for (auto r : res) {
    for (auto i : r) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<db, db>;
using vi = vector<int>;
using vb = vector<bool>;
using vc = vector<char>;
using vl = vector<ll>;
using vd = vector<db>;
using vs = vector<str>;
using vpi = vector<pi>;
using vpl = vector<pl>;
using vpd = vector<pd>;
template <class T>
using V = vector<T>;
const ll MOD = 1e9 + 7;
const int MX = 2e+5;
const ll INF = 1e18;
const db PI = acos((db)-1);
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
constexpr int p2(int x) { return 1 << x; }
constexpr int msk2(int x) { return p2(x) - 1; }
void setIO() {
  cin.tie(0)->sync_with_stdio(0);
  cin.exceptions(cin.failbit);
}
int main() {
  setIO();
  int n;
  cin >> n;
  V<vi> res(n, vi(n));
  for (int i = (0); i < (n / 2); ++i) {
    for (int j = (0); j < (n / 2); ++j) {
      res[i][j] = 2 * n * i + 4 * j;
      res[i][j + n / 2] = res[i][j] + 1;
      res[i + n / 2][j] = res[i][j] + 2;
      res[i + n / 2][j + n / 2] = res[i][j] + 3;
    }
  }
  for (auto r : res) {
    for (auto i : r) cout << i << ' ';
    cout << '\n';
  }
  return 0;
}

```