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 a[200013];
char s[200013];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    long double n;
    scanf("%lf", &n);
    double a, ans = 0;
    long double q = (n * 2 - 2) * 180.0 / (n * 2);
    long double m = 2.0;
    long double x = q / m;
    a = 0;
    int e = n;
    int f = e / 2;
    int g = f + 1;
    for (int i = 0; i < e / 2; i++) {
      long double y = x * 3.14159265358979 / 180.0;
      a += cos(y);
      x = q - (90.0 - x) - 90.0;
    }
    a *= 2.0;
    a += 1;
    a /= 2;
    double f1 = 180.0 / n * f;
    double g1 = 180.0 / n * g;
    f1 = a * sin(f1 / 2 * 3.14159265358979 / 180) * 2;
    g1 = a * sin(g1 / 2 * 3.14159265358979 / 180) * 2;
    ans = (f1 + g1) * (sqrt(2.0) / 2);
    printf("%lf\n", ans);
  }
}