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;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 5;
unsigned long long gcd(unsigned long long a, unsigned long long b) {
  return (b == 0 ? a : gcd(b, a % b));
}
unsigned long long lcm(unsigned long long a, unsigned long long b) {
  return (a * (b / gcd(a, b)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    double ans = 1.0 / sin(1.0l * acos(-1) / (2 * n)) *
                 sin(1.0l * acos(-1) / 4 + (n / 2) * 1.0l * acos(-1) / (2 * n));
    cout << fixed << setprecision(10) << ans << endl;
  }
  return 0;
}