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 collections import defaultdict, deque
import heapq, bisect, math, itertools, string, queue, copy, time

import sys
# sys.setrecursionlimit(10**8)
input = sys.stdin.readline
INF = float('inf')
mod = 10**9 + 7
eps = 10**-7


def inp():
    '''
    一つの整数
    '''
    return int(input())


def inpl():
    '''
    一行に複数の整数
    '''
    return list(map(int, input().split()))


def inpl_str():
    '''
    一行に複数の文字
    '''
    return list(input().split())


t = inp()
ans = []
for _ in range(t):
    n = inp()
    x_2 = 1.0 / (2.0 * (1.0 - math.cos(math.pi / n)))
    r = 2 * math.sqrt(x_2 - 0.5 * 0.5)
    ans.append(r)

print("\n".join([str(i) for i in ans]))