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>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3")
using namespace std;
const long long maxn = 2e5 + 69, inf = 1e9 + 420, lg = 20;
inline void optimizations() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
}
long double inp() {
  long double x;
  cin >> x;
  return x;
}
int main() {
  while (0) {
    long double a;
    cin >> a;
    cout << sin(a) << endl;
  }
  cout << fixed;
  long double t = inp();
  while (t--) {
    long double n = inp();
    long double ang = 1.0 * 360.0 / (4 * n);
    ang = (ang * 3.14159265359) / (180.0);
    cout << setprecision(6) << cos(ang) / sin(ang) << '\n';
  }
}