#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int ret = 0, t = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') t = -1, c = getchar();
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret * t;
}
const int MAXN = 1010;
int a[MAXN * MAXN], ans[MAXN][MAXN];
bool cmp(int x, int y) {
  if ((x & 3) != (y & 3)) return (x & 3) < (y & 3);
  return x < y;
}
int main() {
  int n = read(), i, j;
  for (i = 1; i <= n * n; i++) a[i] = i - 1;
  sort(a + 1, a + n * n + 1, cmp);
  int hn = n >> 1, now = 0;
  for (i = 1; i <= hn; i++)
    for (j = 1; j <= hn; j++) ans[i][j] = a[++now];
  for (i = 1; i <= hn; i++)
    for (j = hn + 1; j <= n; j++) ans[i][j] = a[++now];
  for (i = hn + 1; i <= n; i++)
    for (j = 1; j <= hn; j++) ans[i][j] = a[++now];
  for (i = hn + 1; i <= n; i++)
    for (j = hn + 1; j <= n; j++) ans[i][j] = a[++now];
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) cout << ans[i][j] << ' ';
    cout << endl;
  }
  return 0;
}

### Prompt

Please formulate 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;
inline int read() {
  int ret = 0, t = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') t = -1, c = getchar();
  while (c >= '0' && c <= '9') ret = ret * 10 + c - '0', c = getchar();
  return ret * t;
}
const int MAXN = 1010;
int a[MAXN * MAXN], ans[MAXN][MAXN];
bool cmp(int x, int y) {
  if ((x & 3) != (y & 3)) return (x & 3) < (y & 3);
  return x < y;
}
int main() {
  int n = read(), i, j;
  for (i = 1; i <= n * n; i++) a[i] = i - 1;
  sort(a + 1, a + n * n + 1, cmp);
  int hn = n >> 1, now = 0;
  for (i = 1; i <= hn; i++)
    for (j = 1; j <= hn; j++) ans[i][j] = a[++now];
  for (i = 1; i <= hn; i++)
    for (j = hn + 1; j <= n; j++) ans[i][j] = a[++now];
  for (i = hn + 1; i <= n; i++)
    for (j = 1; j <= hn; j++) ans[i][j] = a[++now];
  for (i = hn + 1; i <= n; i++)
    for (j = hn + 1; j <= n; j++) ans[i][j] = a[++now];
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) cout << ans[i][j] << ' ';
    cout << endl;
  }
  return 0;
}

```