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:
from math import *
from collections import *
from itertools import *
from bisect import *
from heapq import *
from operator import *
from sys import *
setrecursionlimit(1000000)
d=defaultdict(lambda:0,{})
def io():
    return map(int,input().split())
def op():
    return list(map(int,input().split()))
def o():
    return int(input())
def r(x):
    return range(len(x))
def mk(a,b):
    if a>b:return b,a
    else:return a,b
def kl(con,x=0):
    if x==0:print('Yes') if con else print('No')
    elif x==1:print('yes') if con else print('no')
    elif x==2:print('YES') if con else print('NO')
MOD = 1000000007
MAX=float('inf')
MIN=-float('inf')
p=input
for _ in range(o()):
    n=o()
    n*=2
    # l=op()
    y = 1 / (2 * (sin(pi / n)))
    print(sqrt(y ** 2 - (1 / 2) ** 2) * 2)