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 as m

def fu(a):
    return (a/180)*m.pi

t=int(input())
for _ in range(t):
    n=int(input())
    a=n//2
    e=a
    if a%2!=0:
        a=a+1
    b=(180-360/(2*n))/2
    s=0
    for i in range(1,a//2+1):
        s=s+m.sin(fu((2*i-1)*b-(i-1)*180))
    a=(n-a)//2
    u=0
    b=b*2
    for i in range(1,a+1):
        u=u+m.sin(fu(i*b-(2*i-1)*90))
    print((2*s+2*u+1)/m.sqrt(2))