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;
void prt(long x) { cerr << x; }
void prt(long long x) { cerr << x; }
void prt(unsigned x) { cerr << x; }
void prt(unsigned long x) { cerr << x; }
void prt(unsigned long long x) { cerr << x; }
void prt(float x) { cerr << x; }
void prt(double x) { cerr << x; }
void prt(long double x) { cerr << x; }
void prt(char x) { cerr << '\'' << x << '\''; }
void prt(const char *x) { cerr << '\"' << x << '\"'; }
void prt(const string &x) { cerr << '\"' << x << '\"'; }
void prt(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void prt(const pair<T, V> &x) {
  cerr << '{';
  prt(x.first);
  cerr << ',';
  prt(x.second);
  cerr << '}';
}
template <typename T>
void prt(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), prt(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  prt(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void err(istream_iterator<string> it) {}
template <typename T, typename... brgs>
void err(istream_iterator<string> it, T a, brgs... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
long long kadane(long long *a, long long n) {
  long long max_so_far = INT_MIN, max_ending_here = 0;
  for (__typeof(n) i = (0); i != (n); i += 1 - 2 * ((0) > (n))) {
    max_ending_here += a[i];
    max_so_far = max(max_so_far, max_ending_here);
    if (max_ending_here < 0) max_ending_here = 0;
  }
  return max_so_far;
}
void solve() {
  long long n;
  cin >> n;
  double ans = 1.0 / tan(3.1415926535897932384626 / (2 * n));
  cout << fixed;
  cout << setprecision(10) << ans << endl;
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}