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 long long max3 = 1003;
const long long max4 = 10004;
const long long max5 = 100005;
const long long max6 = 1000006;
const long long max7 = 10000007;
const long long lg4 = 13;
const long long lg5 = 17;
const long long lg6 = 20;
const long long INF = 2LL * 1000000000;
const long long INFLL = 9LL * 1000000000 * 1000000000;
const long long M = 1e9 + 7;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
  }
  return res;
}
int n;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("input.txt", "r")) {
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
  }
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long double PI = 3.141592653590;
    long double ans = 1 / (double)tan(PI / (double)(2 * n));
    cout << fixed << setprecision(12) << ans << '\n';
  }
  return 0;
}