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:
# for #!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase


def main():
	n = int(input())
	sup = [[8, 9, 1, 13],
	[3, 12, 7, 5],
	[0, 2, 4, 11],
	[6, 10,15, 14]]
	ans = [[0 for i in range(n)] for j in range(n)]

	# l = []
	m = (n//4)**2
	mx = len(bin(n**2-1)[2:])
	l = [0 for i in range(m)]
	mx2 = len(bin(m-1)[2:])
	for i in range(m):
		a = bin(i)[2:]
		a = (mx2-len(a))*"0"+a
		l[i] = int(a+"0"*(mx-len(a)), 2)
	# # print(l)
	ct = 0
	tct = 0
	i_of = 0
	j_of = 0
	offs = []
	for i in range(n//4):
		for j in range(n//4):
			offs.append((i, j))
	# print(offs, m, n//4)
	for ct in range(m):
		for i in range(4):
			# if i%4==0:
				# ct+=1
			for j in range(4):
				# print(offs[ct][0]*4 + i, offs[ct][1]*4 + j)
				ans[offs[ct][0]*4 + i][offs[ct][1]*4 + j] = sup[i%4][j%4]+l[ct]
		# ct+=1
	for i in range(n):
		for j in range(n):
			print(int(ans[i][j]), end=" ")
		print()
# print(ans)




# region fastio

BUFSIZE = 8192


class FastIO(IOBase):
	newlines = 0

	def __init__(self, file):
		self._fd = file.fileno()
		self.buffer = BytesIO()
		self.writable = "x" in file.mode or "r" not in file.mode
		self.write = self.buffer.write if self.writable else None

	def read(self):
		while True:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			if not b:
				break
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines = 0
		return self.buffer.read()

	def readline(self):
		while self.newlines == 0:
			b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
			self.newlines = b.count(b"\n") + (not b)
			ptr = self.buffer.tell()
			self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
		self.newlines -= 1
		return self.buffer.readline()

	def flush(self):
		if self.writable:
			os.write(self._fd, self.buffer.getvalue())
			self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
	def __init__(self, file):
		self.buffer = FastIO(file)
		self.flush = self.buffer.flush
		self.writable = self.buffer.writable
		self.write = lambda s: self.buffer.write(s.encode("ascii"))
		self.read = lambda: self.buffer.read().decode("ascii")
		self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
	main()