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;
char chaine[200005];
double comp(int n, double angle, double l, double curr_angle) {
  double pi = 3.14159265358979323846;
  double high_x = 0.0;
  double low_x = 0.0;
  double high_y = 0.0;
  double low_y = 0.0;
  for (int i = 0; i < 2 * n; i++) {
    curr_angle += angle;
    double x = cos(curr_angle - (pi / 2));
    double y = sin(curr_angle - (pi / 2));
    high_x = max(high_x, x);
    low_x = min(low_x, x);
    high_y = max(high_y, y);
    low_y = min(low_y, y);
  }
  double res = max(high_x - low_x, high_y - low_y);
  return res;
}
void prob() {
  int n;
  scanf("%d", &n);
  double pi = 3.14159265358979323846;
  double angle = (2 * pi) / (2 * n);
  double l = sqrt(1 / (2 * (1 - cos(angle))));
  double debut = 0.0;
  double fin = (2 * pi) / (4 * n);
  while (fin - debut > 1e-10) {
    double p1 = (debut * 2 + fin) / 3;
    double p2 = (debut + fin * 2) / 3;
    double res1 = comp(n, angle, l, p1);
    double res2 = comp(n, angle, l, p2);
    if (res1 > res2) {
      debut = p1;
    } else {
      fin = p2;
    }
  }
  printf("%.8lf\n", l * comp(n, angle, l, (debut + fin) / 2));
}
int main() {
  int t;
  scanf("%d", &t);
  for (int i = 0; i < t; i++) prob();
  return 0;
}