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 cos, pi
def grad_to_rad(grad):
    return grad/360*pi*2
def mycos(grad):
    return cos(grad_to_rad(grad))
def printv(a, sep = ' '):
    print(sep.join(map(str, a)))
def solve(n):
    k = 1
    N = n*2
    q = n//2 - 2
    ang = (180*(N - 2)/N)
    prev = 90
    for i in range(q + 1):
        cur = ang - prev
        prev = 180 - cur
        k += mycos(cur)*2
    return k
def main():
    t = int(input())
    ans = [0]*t
    for i in range(t):
        ans[i] = solve(int(input()))
    printv(ans, '\n')
main()