#include <bits/stdc++.h>
using namespace std;
template <typename T>
void watch(vector<T> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "Look above\n";
  cout << "\n";
}
const long long INF = 1e15;
const long long MAX = 1e9 + 3;
long long MOD = 998244353;
long long fact(long long n) {
  long long ans = 1;
  for (long long i = 1; i <= n; i++) ans = (ans * i) % MOD;
  return ans;
}
void print_Xor(long long a[][100], long long n) {
  for (long long i = 0; i < n; i++) {
    long long Xor = 0;
    for (long long j = 0; j < n; j++) Xor = (Xor ^ a[i][j]);
    cout << Xor << " ";
  }
  cout << "\n";
  for (long long j = 0; j < n; j++) {
    long long Xor = 0;
    for (long long i = 0; i < n; i++) Xor = (Xor ^ a[i][j]);
    cout << Xor << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  long long ans[n + 2][n + 2];
  long long curr = 0;
  for (long long i = 0; i < n / 2; i++) {
    for (long long j = 0; j < n / 2; j++) {
      ans[i][j] = curr * 4;
      ans[i + n / 2][j] = curr * 4 + 2;
      ans[i + n / 2][j + n / 2] = curr * 4 + 1;
      ans[i][j + n / 2] = curr * 4 + 3;
      curr++;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) cout << ans[i][j] << " ";
    cout << "\n";
  }
}

### Prompt

In CPP, your task is to solve 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;
template <typename T>
void watch(vector<T> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "Look above\n";
  cout << "\n";
}
const long long INF = 1e15;
const long long MAX = 1e9 + 3;
long long MOD = 998244353;
long long fact(long long n) {
  long long ans = 1;
  for (long long i = 1; i <= n; i++) ans = (ans * i) % MOD;
  return ans;
}
void print_Xor(long long a[][100], long long n) {
  for (long long i = 0; i < n; i++) {
    long long Xor = 0;
    for (long long j = 0; j < n; j++) Xor = (Xor ^ a[i][j]);
    cout << Xor << " ";
  }
  cout << "\n";
  for (long long j = 0; j < n; j++) {
    long long Xor = 0;
    for (long long i = 0; i < n; i++) Xor = (Xor ^ a[i][j]);
    cout << Xor << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  long long ans[n + 2][n + 2];
  long long curr = 0;
  for (long long i = 0; i < n / 2; i++) {
    for (long long j = 0; j < n / 2; j++) {
      ans[i][j] = curr * 4;
      ans[i + n / 2][j] = curr * 4 + 2;
      ans[i + n / 2][j + n / 2] = curr * 4 + 1;
      ans[i][j + n / 2] = curr * 4 + 3;
      curr++;
    }
  }
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) cout << ans[i][j] << " ";
    cout << "\n";
  }
}

```