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 double PI = 3.1415926535;
int T;
double n;
int main() {
  scanf("%d", &T);
  while (T--) {
    scanf("%lf", &n);
    double t = ((2.0) / (n * 2));
    double qaq = 0;
    double bc = (0.5) / sin((t / 2) * PI);
    double ans = 0;
    for (int i = 1; i <= n - 1; i++) {
      qaq += t;
      double qwq;
      if (qaq < 0.25)
        qwq = 0.25 - qaq;
      else if (qaq < 0.5)
        qwq = qaq - 0.25;
      else if (qaq < 0.75)
        qwq = 0.75 - qaq;
      else
        qwq = qaq - 0.75;
      ans = max(ans, bc * cos(qwq * PI));
    }
    printf("%.8lf\n", ans * 2);
  }
  return 0;
}