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:
from sys import stdin, stdout
from math import pi, sin, cos

def cal(n, angle, r):
    ang = 180. / n
    res = 0
    for i in range(n):
        res = max(res, abs(sin(angle * pi / 180.)) * r)
        res = max(res, abs(cos(angle * pi / 180.)) * r)

        angle += ang
    return res

for _ in range(int(input())):
    n = int(input())
    angle = 180. / n
    rd = .5 / sin(angle / 2. * pi / 180.)

    l, r = 0, angle
    cnt = 100
    while cnt:
        cnt -= 1

        m1, m2 = (l + l + r) / 3., (l + r + r) / 3.
        t1, t2 = cal(n, m1, rd), cal(n, m2, rd)
        if t1 < t2:
            r = m2
        else:
            l = m1

    ans = cal(n, l, rd)

    print('{:.9f}'.format(ans * 2.))