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 long double PI = acos(-1.0);
long long n;
long double r, alpha, len;
long double f(long long x) {
  alpha = (long double)PI / (2 * n);
  r = (long double)1.0 / (sin(alpha));
  len = r * sin((long double)x * alpha);
  return len;
}
signed main() {
  cout << fixed << setprecision(10);
  long long q;
  cin >> q;
  long double ans, l1, l2;
  while (q--) {
    cin >> n;
    l1 = f(n / 2), l2 = f((n / 2) + 1);
    l1 = sqrt(l1 * l1 / 2), l2 = sqrt(l2 * l2 / 2);
    ans = l1 + l2;
    cout << ans << "\n";
  }
  return 0;
}