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())
 
for i in range(0, n, 4):
    print(' '.join((str(j) for j in range(2*(i//4)*n, 2*(i//4)*n+2*n, 2))))
    print(' '.join((str(j) for j in range(n*n-2*((i+1)//4)*n-1, n*n-2*((i+1)//4)*n-2*n, -2))))
    print(' '.join((str(j) for j in range(2*((i+2)//4)*n+1, 2*((i+2)//4)*n+2*n+1, 2))))
    print(' '.join((str(j) for j in range(n*n-2*((i+3)//4)*n-2, n*n-2*((i+3)//4)*n-2*n-2, -2))))

#Не смотри мой код! На этот раз всё будет о**енно!
 
#for i in range(n):
#    for j in range(n):
#        if i%4==0:
#            print(2*j+2*(i//4)*n, end=' ')
#        elif i%4==1:
#            print(n*n-1-2*j-2*(i//4)*n, end=' ')
#        elif i%4==2:
#            print(1+2*j+2*(i//4)*n, end=' ')
#        elif i%4==3:
#            print(n*n-2-2*j-2*(i//4)*n, end=' ')
#    print('')