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:
"""def power(x, y): 
    res = 1
    x=x
    while (y > 0): 
        if ((y & 1) == 1) : 
            res = (res * x) 
        y = y >> 1
        x = (x * x)
    return res""" 
"""def fact(n):
    if(n==0):
        return 1
    if(n==1):
        return 1
    return fact(n-1)+fact(n-2)"""

"""def sieveOfEratothenes(n):
    dp = [True]*(n+1)
    dp[1]=True
    p=2
    while(p*p<n+1):
        if(dp[p]==True):
            for i in range(p*p, n+1, p):
                dp[i]=False
        p+=1
    return dp"""

import math
#import combinations
 
 
t=int(input())
for _ in range(t):
    n=int(input())
    #s=input()
    #arr=list(map(int,input().strip()))
    #b=list(map(int,input().split()))
    #d={}1
    #flag=0
    n=2*n
    a= math.pi/n
    radius = 1/(2*math.sin(a))
    #ans=radius*2
    ans = 2*radius*math.cos(a)
    print(ans)