#include <bits/stdc++.h>
using namespace std;
const long long INF = 10E16;
const long long mask = (1 << 9) - 1;
const long long base = 100000000000;
const long double eps = 0.0000001;
struct pp {
  long long fi, sc;
  bool const operator<(const pp& b) {
    if (fi == b.fi) return sc < b.sc;
    return fi < b.fi;
  }
  bool const operator>(const pp& b) {
    if (fi == b.fi) return sc > b.sc;
    return fi > b.fi;
  }
  bool const operator==(const pp& b) {
    if (fi == b.fi && sc == b.sc) return 1;
    return 0;
  }
};
bool const operator<(const pp& a, const pp& b) {
  if (a.fi == b.fi) return a.sc < b.sc;
  return a.fi < b.fi;
}
long long n, A[2007][2007];
set<long long> S;
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  long long r = 0;
  for (long long i = 1; i <= n; i += 4) {
    for (long long j = 1; j <= n; j += 4) {
      for (long long ii = i; ii < i + 4; ii++) {
        for (long long jj = j; jj < j + 4; jj++) {
          A[ii][jj] = r;
          r++;
        }
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cout << A[i][j] << ' ';
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create 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;
const long long INF = 10E16;
const long long mask = (1 << 9) - 1;
const long long base = 100000000000;
const long double eps = 0.0000001;
struct pp {
  long long fi, sc;
  bool const operator<(const pp& b) {
    if (fi == b.fi) return sc < b.sc;
    return fi < b.fi;
  }
  bool const operator>(const pp& b) {
    if (fi == b.fi) return sc > b.sc;
    return fi > b.fi;
  }
  bool const operator==(const pp& b) {
    if (fi == b.fi && sc == b.sc) return 1;
    return 0;
  }
};
bool const operator<(const pp& a, const pp& b) {
  if (a.fi == b.fi) return a.sc < b.sc;
  return a.fi < b.fi;
}
long long n, A[2007][2007];
set<long long> S;
int main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  long long r = 0;
  for (long long i = 1; i <= n; i += 4) {
    for (long long j = 1; j <= n; j += 4) {
      for (long long ii = i; ii < i + 4; ii++) {
        for (long long jj = j; jj < j + 4; jj++) {
          A[ii][jj] = r;
          r++;
        }
      }
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      cout << A[i][j] << ' ';
    }
    cout << endl;
  }
  return 0;
}

```