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 = acos(-1);
const double eps = 1e-8;
const int N = 1e5 + 10;
int t, n;
double ans, a;
double cal(double b) {
  double rs = 0;
  for (double i = b; i <= pi / 2; i += a) {
    rs += cos(i);
  }
  return rs;
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    ans = 1e5;
    a = pi / (double)n;
    for (double i = 0, p; i < a; i += 0.00004) {
      p = i;
      while (p < pi / 2 && abs(pi / 2 - p) > eps) p += a;
      p -= pi / 2;
      ans = min(max(cal(i) + cal(a - i), cal(a - p) + cal(p)), ans);
    }
    printf("%.8lf\n", ans);
  }
}