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
def solve():
    n = int(input())
    radperseg = math.pi / 180 * 360 / 2 / n
    l = 0 ; r = math.pi / 2 - radperseg
    prev = -1
    ans = 0
    while l < r:
        ans = (l+r) / 2
        fr = math.sin(radperseg * (n - 1)/2 + r) - math.cos(r)
        fm = math.sin(radperseg * (n - 1)/2 + ans) - math.cos(ans)
        if fm*fr > 0:
            r = ans
        else:
            l = ans
        if abs(fm - prev) < 1e-9:
            break
        prev = fm
    print(2 * math.sqrt(1/(2*(1 - math.cos(radperseg)))) * math.cos(ans))
for i in range(int(input())): solve()