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 sys
from math import pi, sin, cos
input = sys.stdin.readline
for _ in range(int(input())):
  N = int(input())
  l = 1 / sin(pi / N)
  arg = pi * (N - 2) / N
  #print(l, l * cos(pi / 2 - arg / 2))
  x0 = 0
  x1 = 0
  x2 = 0
  x3 = pi / 2 - arg / 2
  while x3 - x0 > pow(0.1, 10):
    x1 = (x0 * 2 + x3) / 3
    x2 = (x0 + x3 * 2) / 3
    r1 = max(cos(x1), cos(pi / 2 + x1))
    r2 = max(cos(x2), cos(pi / 2 + x2))
    if r1 > r2: x0 = x1
    else: x3 = x2
  print(l + cos(x1) * l)