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;
long long int minf = numeric_limits<long long int>::min();
long long int inf = numeric_limits<long long int>::max();
const long long int N = 500007;
const long long int MOD = 100000007;
void sol() {
  double t = 0, n = 0, a = 0, c = 0, d = 0, k = 0, f = 0, m = 0, r = 0;
  double b;
  cin >> n;
  n = n * 2;
  a = 360 / n;
  b = a;
  m = 90;
  while (abs(m - b) > 1e-9) {
    d += (2.0 * sin((b * 3.141592654) / 180));
    b += a;
  }
  d = d + 1;
  cout << setprecision(10) << d;
}
int main() {
  long long int t = 1;
  cin >> t;
  while (t--) {
    sol();
    cout << "\n";
    ;
  }
  return 0;
}