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:
t=int(input())
from math import pi,sin,cos
for i in range(t):
    n=int(input())
    angle=((2*n-2)*pi)/(2*n)
    l=[angle/2]
    a=angle/2
    k=n//2
    for j in range(n//2-1):
            l.append(angle-pi+a)
            a=l[-1]
            if pi/2-l[-1]<=pi/4:
                k=len(l)+1
    s1,s2,s3=0,0,0
    for j in range(max(1,k-1)):
        s1+=sin(l[j])
    for j in range(max(1,k-1),n//2):
        s2+=2*cos(l[j])
        s3+=cos(l[j])
    a=1+s2+2*s1
    s3+=0.5
    b=a/(2**0.5)
    print(b)