#include <bits/stdc++.h>
using namespace std;
void print(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
long long int power1(long long int x) {
  long long int u = 0;
  while (x > 0) {
    u++;
    x = x / 2;
  }
  return u;
}
long long int powM(long long int x, long long int y, long long int m) {
  if (m == 1) {
    return 0;
  }
  long long int ans = 1, r = 1;
  x %= m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= m;
    }
    r <<= 1;
    x *= x;
    x %= m;
  }
  return ans;
}
long long int powm(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t2 = 1;
  long long int M = 1e9 + 7;
  while (t2--) {
    long long int n;
    cin >> n;
    long long int a[n][n];
    long long int r = n / 4;
    long long int p = 0;
    long long int k = 0;
    for (long long int i = 0; i < n; i++) {
      p = (4 * (i % 4)) + (i / 4) * 16 * r;
      k = -1;
      for (long long int j = 0; j < n; j++) {
        if (j % 4 == 0) {
          k++;
          p = (4 * (i % 4)) + (i / 4) * 16 * r;
        }
        cout << p + 16 * k << " ";
        p++;
      }
      cout << "\n";
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
void print(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
long long int power1(long long int x) {
  long long int u = 0;
  while (x > 0) {
    u++;
    x = x / 2;
  }
  return u;
}
long long int powM(long long int x, long long int y, long long int m) {
  if (m == 1) {
    return 0;
  }
  long long int ans = 1, r = 1;
  x %= m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= m;
    }
    r <<= 1;
    x *= x;
    x %= m;
  }
  return ans;
}
long long int powm(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t2 = 1;
  long long int M = 1e9 + 7;
  while (t2--) {
    long long int n;
    cin >> n;
    long long int a[n][n];
    long long int r = n / 4;
    long long int p = 0;
    long long int k = 0;
    for (long long int i = 0; i < n; i++) {
      p = (4 * (i % 4)) + (i / 4) * 16 * r;
      k = -1;
      for (long long int j = 0; j < n; j++) {
        if (j % 4 == 0) {
          k++;
          p = (4 * (i % 4)) + (i / 4) * 16 * r;
        }
        cout << p + 16 * k << " ";
        p++;
      }
      cout << "\n";
    }
  }
  return 0;
}

```