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:
# -*- coding: utf-8 -*-
import sys
from collections import defaultdict, deque
from math import sqrt, gcd, factorial, tan, pi, sin, cos
def input(): return sys.stdin.readline()[:-1] # warning not \n
# def input(): return sys.stdin.buffer.readline()[:-1]



def solve():
    n = int(input())
    n *= 2
    # m = n - 1
    # S = m / (4 / tan(pi/m)

    # R = 1 / (2 * sin(pi/m))
    # r = R * cos(pi/m)

    # c = sqrt(r)
    # b = sqrt(1 / 2)

    if n == 2:
        print(1)
    else:
        print(1/tan(pi / n))



t = 1
t = int(input())
for case in range(1,t+1):
    ans = solve()



"""





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