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;
template <typename T>
T Sqr(T x) {
  T n = x * x;
  return n;
}
template <typename T>
T Pow(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * Pow(B, P - 1);
  else
    return Sqr(Pow(B, P / 2));
}
template <typename T>
T Abs(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
T Gcd(T a, T b) {
  if (a < 0) return Gcd(-a, b);
  if (b < 0) return Gcd(a, -b);
  return (b == 0) ? a : Gcd(b, a % b);
}
template <typename T>
T Lcm(T a, T b) {
  if (a < 0) return Lcm(-a, b);
  if (b < 0) return Lcm(a, -b);
  return a * (b / Gcd(a, b));
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    double n, theta, ans;
    cin >> n;
    theta = acos(-1) / (2 * n);
    ans = 1 / tan(theta);
    printf("%.9lf\n", ans);
  }
  return 0;
}