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:
"""
arr = list(map(int, input().split()))
n,k=map(int, input().split())
"""
import math
import sys
# input = sys.stdin.readline

############ ---- Input Functions ---- ############
def inp():
    return(int(input()))
def inlt():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))

test_cases = int(input())
for _ in range(test_cases):
    sides = int(input())
    sides *= 2
    apothem = 1 / (2 * math.tan((180 / sides) * (math.pi/180)))   
    print(2 * apothem)
# for _ in range(test_cases):
#     size = int(input())
#     arr = inlt()
#     maxx = -float('inf')
#     temp = []
#     max_diff = 0
#     #Checks the maximum number and difference of decreasing numbers, the moment it increases again, it rechecks for a bigger difference
#     for i in range(size):
#         if arr[i] < maxx:
#             max_diff = max(max_diff, maxx - arr[i])
#         maxx = max(arr[i], maxx)
#     i = 0
#     index = 0
#     while i < max_diff:
#         i += 2 ** index
#         index += 1
#     print(index)