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 dx[] = {0, 0, +1, -1};
int dy[] = {+1, -1, 0, 0};
double value(double n) { return (n * (3.1416 / 180.00)); }
int main() {
  int tc;
  cin >> tc;
  while (tc--) {
    double n, kona, angle, b, x, ans = 1.0, a;
    cin >> n;
    kona = int(n - 1) / 2;
    angle = (180.0 / n) * kona;
    b = sqrt(1.0 / (2.0 * (1.0 - cos(value(180.0 / n)))));
    x = sqrt(2.00 * b * b * (1 - cos(value(angle))));
    a = sqrt((x * x) / 2.00);
    ans += (2.0 * a);
    printf("%.15lf\n", ans);
  }
  return 0;
}