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", "inline", "-ffast-math")
#pragma GCC target("avx,sse2,sse3,sse4,mmx")
using namespace std;
using ll = long long;
const int N = 2e5 + 10;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int t, n;
  cin >> t;
  while (t--) {
    cin >> n;
    n *= 2;
    double ans = 0, k, fk;
    k = 180.0 * (n - 2) / n;
    fk = 180.0 - k;
    for (int i = 2; i <= (n >> 2); ++i) {
      ans += sin((k - 90) * 3.14159265 / 180.0);
      k -= fk;
    }
    printf("%.9f\n", ans * 2 + 1);
  }
  return 0;
}