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("O3")
using namespace std;
const int N = 1e6 + 5, LG = 18;
const double pi = acos(-1);
const long long mod = 1e9 + 7;
mt19937_64 rnd(239);
long long power(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x %= p;
  while (y) {
    if (y & 1) res = (res * x) % p;
    y >>= 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    double ans = 1.0 / tan(pi / (2.0 * n));
    cout << fixed << setprecision(10) << ans << "\n";
  }
  return 0;
}