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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    double z = 1.0 / (2.0 - 2.0 * cos(acos(-1.0) / (n * 1.0)));
    z = sqrt(z);
    double x1 = sin(acos(-1.0) / (n * 1.0) * round(n / 4.0)) * z * sqrt(2.0);
    double x2 =
        sin(acos(-1.0) / 2.0 - (acos(-1.0) / (n * 1.0) * round(n / 4.0))) * z *
        sqrt(2.0);
    x1 += x2;
    printf("%.9lf\n", x1);
  }
  return 0;
}