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.
#include <bits/stdc++.h>
using namespace std;
int T;
int n;
int a[1111][1111];
pair<int, int> b[17];
void check() {
  for (int i = 0; i < n; i++) {
    int val = 0;
    for (int j = 0; j < n; j++) {
      val ^= a[i][j];
    }
    printf("%d row : %d\n", i, val);
    val = 0;
    for (int j = 0; j < n; j++) {
      val ^= a[j][i];
    }
    printf("%d col : %d\n", i, val);
  }
}
int main() {
  b[0] = pair<int, int>(2, 0);
  b[1] = pair<int, int>(0, 2);
  b[2] = pair<int, int>(2, 1);
  b[3] = pair<int, int>(1, 0);
  b[4] = pair<int, int>(2, 2);
  b[5] = pair<int, int>(1, 3);
  b[6] = pair<int, int>(3, 0);
  b[7] = pair<int, int>(1, 2);
  b[8] = pair<int, int>(0, 0);
  b[9] = pair<int, int>(0, 1);
  b[10] = pair<int, int>(3, 1);
  b[11] = pair<int, int>(2, 3);
  b[12] = pair<int, int>(1, 1);
  b[13] = pair<int, int>(0, 3);
  b[14] = pair<int, int>(3, 3);
  b[15] = pair<int, int>(3, 2);
  scanf("%d", &n);
  int nsq = n * n;
  for (int i = 0; i < 16; i++) {
    for (int j = 0; j < n / 4; j++) {
      for (int k = 0; k < n / 4; k++) {
        a[j * 4 + b[i].first][k * 4 + b[i].second] = j * (n * 4) + k * (16) + i;
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      printf("%d ", a[i][j]);
    }
    printf("\n");
  }
  return 0;
}
