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 sys
import math
pi=math.pi
q=int(input())
for i in range(q):
    a=int(sys.stdin.readline())
    if a//2%2==0:
        n1=a//2+1
    else:
        n1=a//2
    n2=(a-n1)
    rad=1/(2*math.sin(pi/(2*a)))
    b=pow(2*rad*rad-2*rad*rad*math.cos(2*pi/(2*a)*n1),1/2)/pow(2,1/2)
    c=pow(2*rad*rad-2*rad*rad*math.cos(2*pi/(2*a)*n2),1/2)/pow(2,1/2)
    print(b+c)
'''
    1/(2*math.sin(pi/(2*a)))
    
    print( )
           
    print(1/math.tan(pi/(2*a)))
'''