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:
#!/usr/bin/env python3
# -*- coding: utf-8 -*-

import array
from bisect import *
from collections import *
import fractions
import heapq
from itertools import *
import math
import random
import re
import string
import sys

T = int(input())
for t in range(T):
    n = int(input())

    theta = math.pi / (2.0 * n)

    r = 1.0 / (2.0 * math.sin(theta))

    max_x = 0
    for i in range(n):
        # 45deg + 2 * i * theta
        theta2 = (math.pi / 4.0) + 2 * i * theta
        x = r * math.sin(theta2)
        max_x = max(max_x, abs(x))

    print("{:.20f}".format(max_x * 2))