#include <bits/stdc++.h>
using namespace std;
const int Size = 1024;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 15;
int a[5][5] = {{0, 0, 0, 0, 0},
               {0, 0, 1, 2, 3},
               {0, 7, 6, 5, 4},
               {0, 8, 9, 10, 11},
               {0, 15, 14, 13, 12}};
int b[1015][1015];
int n;
void run(int k) {
  int tmp = k / 4;
  for (int i = 1; n >= i; i++) {
    for (int j = 1; n >= j; j++) {
      int x = i % 4, y = j % 4;
      if (x == 0) x = 4;
      if (y == 0) y = 4;
      b[i][j] =
          a[x][y] +
          ((i / 4 + (i % 4 != 0) - 1) * tmp + (j / 4 + (j % 4 != 0)) - 1) * 16;
    }
  }
}
int main() {
  scanf("%d", &n);
  run(n);
  for (int i = 1; n >= i; i++) {
    int tt = 0;
    for (int j = 1; n >= j; j++) {
      if (tt++ > 0) printf(" ");
      printf("%d", b[i][j]);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int Size = 1024;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int maxn = 2e5 + 15;
int a[5][5] = {{0, 0, 0, 0, 0},
               {0, 0, 1, 2, 3},
               {0, 7, 6, 5, 4},
               {0, 8, 9, 10, 11},
               {0, 15, 14, 13, 12}};
int b[1015][1015];
int n;
void run(int k) {
  int tmp = k / 4;
  for (int i = 1; n >= i; i++) {
    for (int j = 1; n >= j; j++) {
      int x = i % 4, y = j % 4;
      if (x == 0) x = 4;
      if (y == 0) y = 4;
      b[i][j] =
          a[x][y] +
          ((i / 4 + (i % 4 != 0) - 1) * tmp + (j / 4 + (j % 4 != 0)) - 1) * 16;
    }
  }
}
int main() {
  scanf("%d", &n);
  run(n);
  for (int i = 1; n >= i; i++) {
    int tt = 0;
    for (int j = 1; n >= j; j++) {
      if (tt++ > 0) printf(" ");
      printf("%d", b[i][j]);
    }
    printf("\n");
  }
  return 0;
}

```