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;
int n;
const double pi = 3.1415926535897932;
void solve() {
  scanf("%d", &n);
  double nj = pi - pi / (2 * n);
  vector<double> lens;
  lens.push_back(1);
  int jn = n / 2;
  double jj = pi / 2 / jn;
  double ns = 0;
  for (int i = 1; i < jn; i++) {
    ns += jj;
    lens.push_back(cos(ns));
  }
  reverse(lens.begin(), lens.end());
  for (int i = (int)lens.size() - 2; i >= 0; i--) {
    lens.push_back(lens[i]);
  }
  double ans = 0;
  for (int i = 0; i < lens.size(); i++) {
    ans += lens[i];
  }
  printf("%.10lf\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    solve();
  }
  return 0;
}