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
from math import ceil,sqrt,sin,pi,tan
from collections import defaultdict
read=sys.stdin.buffer.readline
import heapq
mi=lambda:map(int,read().split())
li=lambda:list(mi())
cin=lambda:int(read())

def isp(n):
    i=2
    while i*i<=n:
        if n%i==0:
            return False
        i+=1
    return True


primes=[]

no_of_chars = 256
def ps(sm):
    if sm<0:
        return 1
    sq=int(sqrt(sm))
    if sq*sq==sm:
        return 0
    return 1





for _t in range(cin()):
    n=cin()
    n=2*n
    ans=1/(2*(tan(pi/n)))
    print(2*ans)