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
import sys
input = sys.stdin.readline


t = int(input())
for _ in range(t):
    n = int(input())
    theta = 2 * n
    y = 1 / math.sin(math.radians(360 / 4 / n)) / 2
    p = [(0, y)]
    
    rot45 =  [math.cos(math.radians(45)), -math.sin(math.radians(45))], [math.sin(math.radians(45)), math.cos(math.radians(45))]
    tmp = p[-1]
    x = rot45[0][0] * tmp[0] + rot45[0][1] * tmp[1]
    y = rot45[1][0] * tmp[0] + rot45[1][1] * tmp[1]
    p[0] = (x, y)
    the = 360 / (2 * n) 
    rot = [math.cos(math.radians(the)), -math.sin(math.radians(the))], [math.sin(math.radians(the)), math.cos(math.radians(the))]
    max_x = 0
    max_y = 0
    for i in range(2 * n - 1):
        tmp = p[-1]
        x = rot[0][0] * tmp[0] + rot[0][1] * tmp[1]
        y = rot[1][0] * tmp[0] + rot[1][1] * tmp[1]
        max_x = max(abs(x), max_x)
        max_y = max(abs(y), max_y)
        p.append((x, y))
    print(2 * max_x)