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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class R>
inline void Readin(R &K) {
  bool F = K = 0;
  char C = getchar();
  while (C < '0' || C > '9') {
    if (C == '-') F = 1;
    C = getchar();
  }
  while (C <= '9' && C >= '0') K = (K << 1) + (K << 3) + C - '0', C = getchar();
  if (F) K = -K;
}
template <class R>
inline bool Checkmax(R &A, R B) {
  return A < B ? A = B, 1 : 0;
}
template <class R>
inline bool Checkmin(R &A, R B) {
  return A > B ? A = B, 1 : 0;
}
int N;
int Ans[1005][1005], Cnt;
int main() {
  Readin(N);
  int N_ = N;
  while (N_) {
    register int R = 1;
    while ((R << 1) <= N_) R <<= 1;
    for (register int i = 0; ++i <= N;)
      for (register int j = N - N_; ++j <= N - N_ + R;) Ans[i][j] = Cnt++;
    N_ -= R;
  }
  for (register int i = 0; ++i <= N;) {
    for (register int j = 0; ++j <= N;) printf("%d ", Ans[i][j]);
    printf("\n");
  }
  return 0;
}