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 math
def rad(angle) :
	return (angle / 180) * math.pi
def dist(a, b, c, d) :
	return math.sqrt((a - c) * (a - c) + (b - d) * (b - d))
tt = int(input())
while tt > 0 :
	tt -= 1
	n = int(input())
	angle = rad(360 / (2 * n))
	l1, l2 = n // 2, n - n // 2
	px, py = 0, 0
	vx, vy = 1, 0
	ans = 0
	cur = 0
	for i in range(1, n + 1) :
		px += vx
		py += vy
		if i == l1 or i == l2 :
			ans += dist(0, 0, px, py)
		cur += angle
		vx = math.cos(cur)
		vy = math.sin(cur)
	print(ans / math.sqrt(2))