#include <bits/stdc++.h>
using namespace std;
inline long long rint() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
inline long long gcd(long long a, long long b) {
  return (b) ? gcd(b, a % b) : a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long pow(long long a, long long b, long long q) {
  long long rtn = 1;
  while (b) {
    if (b & 1) rtn = rtn * a % q;
    a = a * a % q;
    b >>= 1;
  }
  return rtn;
}
inline long long mysqrt(long long n) {
  long long x = sqrt(n);
  while (x * x < n) x++;
  while (x * x > n) x--;
  return x;
}
const long long mod = 1e9 + 7;
int res[1005][1005];
int main() {
  int n = rint() / 4;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int a = i * 4; a < i * 4 + 4; a++) {
        for (int b = j * 4; b < j * 4 + 4; b++) {
          res[a][b] = cnt++;
        }
      }
    }
  }
  for (int i = 0; i < n * 4; i++) {
    for (int j = 0; j < n * 4; j++) cout << res[i][j] << " ";
    cout << "\n";
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 long long rint() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
inline long long gcd(long long a, long long b) {
  return (b) ? gcd(b, a % b) : a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long pow(long long a, long long b, long long q) {
  long long rtn = 1;
  while (b) {
    if (b & 1) rtn = rtn * a % q;
    a = a * a % q;
    b >>= 1;
  }
  return rtn;
}
inline long long mysqrt(long long n) {
  long long x = sqrt(n);
  while (x * x < n) x++;
  while (x * x > n) x--;
  return x;
}
const long long mod = 1e9 + 7;
int res[1005][1005];
int main() {
  int n = rint() / 4;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int a = i * 4; a < i * 4 + 4; a++) {
        for (int b = j * 4; b < j * 4 + 4; b++) {
          res[a][b] = cnt++;
        }
      }
    }
  }
  for (int i = 0; i < n * 4; i++) {
    for (int j = 0; j < n * 4; j++) cout << res[i][j] << " ";
    cout << "\n";
  }
}

```