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

t=int(input())
for _ in range(t):
  n=int(input())
  a=2*(1/math.sin(math.pi/n)*math.sin((math.pi*(n-1))/(2*n)))
  ans=10**18
  l=0
  r=math.pi/(2*n)
  for _ in range(100):
    theta1=(2*l+r)/3
    theta2=(l+2*r)/3
    tmp1=max(a*math.cos(theta1),a*math.sin((math.pi*(n-1))/(2*n)+theta1))
    tmp2=max(a*math.cos(theta2),a*math.sin((math.pi*(n-1))/(2*n)+theta2))
    if tmp1>tmp2:
      l=theta1
    else:
      r=theta2
  print(a*math.cos(r))