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 power(long long a, long long b, long long m = 1000000007) {
  long long ans = 1;
  a = a % m;
  while (b > 0) {
    if (b & 1) ans = (1ll * a * ans) % m;
    b >>= 1;
    a = (1ll * a * a) % m;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
bool cmp(pair<string, int> p1, pair<string, int> p2) {
  return p1.second > p2.second;
}
const int MAXN = 5 * 100000 + 10;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    n *= 2;
    double x = (double)360 / n;
    double y = (double)(180.0 - x) / 2;
    double r = (double)y * ((double)M_PI / 180.0);
    double result = tan(r);
    cout << fixed << setprecision(10) << result << "\n";
  };
}