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 PYTHON3):

Solution:
n = int(input())
a = [[0] * n for _ in range(n)]
def solve(s, xo, yo, vo):
    if s <= 4:
        u = [
            list(map(int, '8 9 1 13'.split())),
            list(map(int, '3 12 7 5'.split())),
            list(map(int, '0 2 4 11'.split())),
            list(map(int, '6 10 15 14'.split()))
        ]
        for i in range(4):
            for j in range(4):
                a[xo + i][yo + j] = u[i][j] + vo
        return
    bs = s // 4
    for i in range(bs):
        for j in range(bs):
            solve(4, xo + i * 4, yo + j * 4, 16 * (i * bs + j))
solve(n, 0, 0, 0)
for i in a:
    print(*i)