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>
template <typename T>
void rd(T& x) {
  long long f = 0, c;
  while ((c = getchar()) < 48 || 57 < c) f ^= !(c ^ 45);
  x = (c & 15);
  while (47 < (c = getchar()) && c < 58) x = x * 10 + (c & 15);
  if (f) x = -x;
}
template <typename T>
inline void pt(T x) {
  if (x < 0) x = -x, putchar('-');
  if (x > 9) pt(x / 10);
  putchar(x % 10 + 48);
}
template <typename T>
inline void pt(T x, long long c) {
  pt(x), putchar(c);
}
template <typename T>
inline T max(const T& x, const T& y) {
  return x < y ? y : x;
}
template <typename T>
inline T min(const T& x, const T& y) {
  return x < y ? x : y;
}
const long long N = 1005;
long long n, a[N][N];
signed main() {
  rd(n);
  long long ind = 0;
  for (long long i = 1; i <= n; i += 4)
    for (long long j = 1; j <= n; j += 4) {
      for (long long x = (i); x <= (i + 3); ++x)
        for (long long y = (j); y <= (j + 3); ++y) a[x][y] = ind++;
    }
  for (long long i = (1); i <= (n); ++i)
    for (long long j = (1); j <= (n); ++j) pt(a[i][j], j == n ? '\n' : ' ');
  return 0;
}