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;
using pii = pair<int, int>;
using ll = long long;
const double PI = 3.14159265359;
long double sq(long double x) { return x * x; }
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    int gon = 2 * n;
    int quart = n / 2 - 1;
    long double angledeg = ((gon - 2) * 180.00) / gon;
    long double anglerad = PI - (angledeg * PI / 180);
    long double currx = 0.0;
    long double curry = 0.0;
    long double currangle = 0.0;
    while (quart--) {
      double mul = 1.0;
      currx += cos(currangle) * mul;
      curry += sin(currangle) * mul;
      currangle += anglerad;
    }
    double lensq = sq(currx) + sq(curry);
    double half = sqrt(lensq / 2) + 0.5;
    printf("%.10lf\n", 2 * half);
  }
  return 0;
}