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:
# c_easy_1.py
#2n-gon problem

import math
import sys
def input():
	return sys.stdin.readline().rstrip()

t = int(input())
answers = []
for i in range(t):
	k = int(input())
	n = 2*k
	x, y = 0,0
	to_do = int(((n-2)/2))
	theta = math.pi - ((n-2)*math.pi)/n
	for i in range(1, to_do+1):
		x += math.cos(i*theta)
		y += math.sin(i*theta)

	ans = math.sqrt(x**2 + y**2)
	answers.append(ans)
print(*answers, sep = '\n')