Problem: The statement of this problem is the same as the statement of problem C2. The only difference is that, in problem C1, n is always even, and in C2, n is always odd.

You are given a regular polygon with 2 ⋅ n vertices (it's convex and has equal sides and equal angles) and all its sides have length 1. Let's name it as 2n-gon.

Your task is to find the square of the minimum size such that you can embed 2n-gon in the square. Embedding 2n-gon in the square means that you need to place 2n-gon in the square in such way that each point which lies inside or on a border of 2n-gon should also lie inside or on a border of the square.

You can rotate 2n-gon and/or the square.

Input

The first line contains a single integer T (1 ≤ T ≤ 200) — the number of test cases.

Next T lines contain descriptions of test cases — one per line. Each line contains single even integer n (2 ≤ n ≤ 200). Don't forget you need to embed 2n-gon, not an n-gon.

Output

Print T real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed 10^{-6}.

Example

Input


3
2
4
200


Output


1.000000000
2.414213562
127.321336469

Solution (in PYTHON3):

Solution:
import sys
import math
from collections import deque
#input = sys.stdin.buffer.readline


def calc(f,x):
	tmp = ang/2+f-math.pi/2
	return math.cos(tmp)*x
for _ in range(int(input())):
	n = int(input())

	ang = (math.pi - math.pi/n)
	toRem = math.pi/2

	l = 0
	for i in range((2*n-1)//4):
		a1 = ang - toRem
		a2 = math.pi/2 - a1

		l += math.sin(a1)
		toRem = a2+math.pi/2

	x = 2*l + 1
	ll = l

	z = (n//2)*math.pi/n
	print(math.cos((math.pi*2-math.pi/2-z)/2 - math.pi/2)*x)
	# r = math.pi/2
	# l = 0
	# for i in range(300):
	# 	print(l,r)
	# 	# ml = (r+2*l)/3
	# 	# mr = (2*r+l)/3

	# 	m = (l+r/2)


	# 	if calc(ml,x)<calc(mr,x):
	# 		r = mr
	# 	else:
	# 		l = ml


	#print(calc(l,x))