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:
from collections import defaultdict as dc
from collections import Counter 
from heapq import *
import math
from bisect import bisect_left,bisect
#bisect gives x and p[x] is element greater than it and out of bound for last one 
#p[x-1] gives equal or smaller and no error for any element.
import sys
from collections import deque as dq
from heapq import heapify,heappush,heappop
mod=10**9 +7
def sinp():
    p=str(sys,stdin.readline())
    return p
def seive(n):
    p=[0]*(n+1)
    for i in range(2,n):
        if p[i]==0:
            for j in range(2*i,n+1,i):
                p[i]=1
    return p
def lower_bound(a,i,n):
    #Number of element is equal or lower than it
    #return bisect(a,i)
    #Number of elementt lower than it
    return bisect_left(a,i)
def upper_bound(a,i,n):
    #Number of element greater than it
    #return n-bisect(a,i)
    #Number of element greater than or equal to it
    return n-bisect_left(a,i)
def inp(): 
    p=int(sys.stdin.readline())
    return p
def line():
    p=list(map(int,sys.stdin.readline().split()))
    return p
def read_mat():
    n=inp()
    a=[]
    for i in range(n):
        a.append(line())
    return a
def digit(n):
    s=str(n)
    p=0
    for i in s:
        p+=(int(i))**2
    return p

def solve(n):
    n=n*2
    return math.tan(math.pi*(0.5* (n-2)/n))




    
t=inp()
for _ in range(t):
    n=inp()
    print(solve(n))