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;
using ll = long long;
using vl = vector<ll>;
using vvl = vector<vl>;
using vd = vector<double>;
const ll delta = 0.0000001;
const ll INF = 1e18 + 7;
const ll MOD = 1e9 + 7;
const double pi = 3.14159265;
ll t;
ll n;
int main() {
  cin.tie(0);
  cout.tie(0);
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cout.precision(15);
  cin >> t;
  for (ll(rock) = 0; (rock) < (t); ++(rock)) {
    cin >> n;
    n *= 4;
    double deg0 = (n - 2) * 180 / (double)n;
    double alfa = deg0 / 2;
    double sin_alfa = sin(alfa * pi / 180);
    double s = 1 / (2 * sin_alfa);
    double deg = 360 / (double)(n);
    double sin_a = sin(deg * pi / 180);
    double b = 90 - deg / 2;
    double sin_b = sin(b * pi / 180);
    double R = sin_b * s / sin_a;
    double h = R * R * sin_a / s;
    cout << 2 * h << endl;
  }
  return 0;
}