#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const double pie = 3.141592653589793;
template <typename T>
T inf() {
  return (T)__builtin_inf();
}
template <typename T>
T gcd(T a, T b) {
  if (b > a) return gcd(b, a);
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
T fast_power(T x, T y, long long m = MOD) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1LL;
    x = (x * x) % m;
  }
  return ans % m;
}
string IntToString(long long a) {
  ostringstream temp;
  temp << a;
  return temp.str();
}
long long findMMI_fermat(long long n, long long M) {
  long long ans = fast_power(n, M - 2, M);
  return ans;
}
long long add(long long a, long long b, long long M) {
  return (((((a) % (M)) + (M)) % (M)) + ((((b) % (M)) + (M)) % (M))) % M;
}
long long sub(long long a, long long b, long long M) {
  return (((((a) % (M)) + (M)) % (M)) + M - ((((b) % (M)) + (M)) % (M))) % M;
}
long long mult(long long a, long long b, long long M) {
  return (((((a) % (M)) + (M)) % (M)) * ((((b) % (M)) + (M)) % (M))) % M;
}
bool isprime(long long a) {
  if (a == 2) {
    return 1;
  }
  if (!(a & 1)) {
    return 0;
  }
  for (long long i = 3; i * i <= a; i += 2) {
    if (a % i == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, i, j;
  cin >> n;
  long long int a[n][n];
  long long int k = 1;
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j < n; j++) {
      a[i][j] = k;
      k += 2;
    }
  }
  k = 0;
  for (i = (n / 2); i < n; i++) {
    for (j = 0; j < n; j++) {
      a[i][j] = k;
      k += 2;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cout << a[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const long long MOD = 1e9 + 7;
const double pie = 3.141592653589793;
template <typename T>
T inf() {
  return (T)__builtin_inf();
}
template <typename T>
T gcd(T a, T b) {
  if (b > a) return gcd(b, a);
  return b == 0 ? a : gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
template <typename T>
T fast_power(T x, T y, long long m = MOD) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1LL;
    x = (x * x) % m;
  }
  return ans % m;
}
string IntToString(long long a) {
  ostringstream temp;
  temp << a;
  return temp.str();
}
long long findMMI_fermat(long long n, long long M) {
  long long ans = fast_power(n, M - 2, M);
  return ans;
}
long long add(long long a, long long b, long long M) {
  return (((((a) % (M)) + (M)) % (M)) + ((((b) % (M)) + (M)) % (M))) % M;
}
long long sub(long long a, long long b, long long M) {
  return (((((a) % (M)) + (M)) % (M)) + M - ((((b) % (M)) + (M)) % (M))) % M;
}
long long mult(long long a, long long b, long long M) {
  return (((((a) % (M)) + (M)) % (M)) * ((((b) % (M)) + (M)) % (M))) % M;
}
bool isprime(long long a) {
  if (a == 2) {
    return 1;
  }
  if (!(a & 1)) {
    return 0;
  }
  for (long long i = 3; i * i <= a; i += 2) {
    if (a % i == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, i, j;
  cin >> n;
  long long int a[n][n];
  long long int k = 1;
  for (i = 0; i < n / 2; i++) {
    for (j = 0; j < n; j++) {
      a[i][j] = k;
      k += 2;
    }
  }
  k = 0;
  for (i = (n / 2); i < n; i++) {
    for (j = 0; j < n; j++) {
      a[i][j] = k;
      k += 2;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cout << a[i][j] << " ";
    }
    cout << "\n";
  }
  return 0;
}

```