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:
# cook your dish here
import math as mt 
  
# Function to find the diagonal  
# of a regular polygon 
def polydiagonal(n, a):  
  
    # Side and side length cannot  
    # be negative  
    if (a < 0 and n < 0):  
        return -1
  
    # diagonal degree converted to radians  
    return  a * mt.cos((180) / n * 3.141592653589793238462643383/ 180)/mt.sin((180) / n * 3.141592653589793238462643383/ 180)
for i in range(int(input())):
    n=int(input())
    print(polydiagonal((2*n),1))