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 funk(n):
    #polygon with n ecken, kleiner als kreis immer hmm
    #die eine flache ecke auf den boden dann die höhe der hälfte der winkel.
    #höchster punkt ist n//2 und der ist wie hoch? naja die höhe aller kanten im winkel in dem sie kommen
    #also erster ist im winkel 0 also null, zweiter ist im winkel letzeter winkel +
    sum = 0
    for i in range(1, int(n//2) +1):
        sum += math.sin(math.radians(i*360/n))
    return sum


count = int(input())
input_l = []
for i in range(count):
    tmp = input()
    input_l.append(int(tmp))

for i in input_l:
    print(funk(2*i))