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;
const int N = (int)1e6 + 7;
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e18 + 7;
long double dp[500], goc[500];
long double Sin(long double i) { return sin(i * 3.14159265 / 180); }
long double Cos(long double i) { return Sin(90 - i); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  dp[2] = 1.0;
  goc[2] = 360;
  for (int i = 3; i <= 200; i++) {
    goc[i] = goc[i - 1] + 360;
    long double n = i * 2;
    long double u = goc[i] / n;
    long double x = u - 90;
    long double z = (180 - u) * (i - 3) / 2;
    long double y = u - z - 90 + x;
    long double a = Cos(x);
    long double b = Sin(x);
    long double f = b / Cos(y);
    long double e = sqrt(f * f - b * b);
    dp[i] = e + a;
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << setprecision(6) << fixed << dp[n];
    cout << "\n";
  }
}