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;
template <typename T, typename U>
static inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T &x, U y) {
  if (y > x) x = y;
}
const long long int mod = (1000000007);
const long long inf = (long long int)1e18 + 17;
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
void solve() {
  double n, ca, sa;
  cin >> n;
  ca = 3.14159265358979323846 / (n);
  sa = 3.14159265358979323846 - ca;
  double cena = ca / 2, l = 0.5, la = sa - (3.14159265358979323846 / 2),
         ans[200];
  for (int i = 1; i <= n / 2; i++) {
    ans[i] = l * (1 + (1 / tan(cena))) * sqrt(2);
    cena += ca;
    l += sin(la);
    la = (3.14159265358979323846 / 2) - la;
    la = sa - (3.14159265358979323846 / 2) - la;
  }
  cena = ca / 2;
  double rans = cos(sa / 2) + sin(sa / 2) / tan(ca / 2);
  rans *= sqrt(2);
  for (int i = 1; i <= n / 2; i++) {
    if (sa - 2 * cena <= (3.14159265358979323846 / 2) &&
        sa + 2 * cena <= 3 * 3.14159265358979323846 / 2) {
      amin(rans, ans[i]);
    }
    cena += ca;
  }
  cout << fixed;
  cout << setprecision(9) << rans << endl;
}
int main() {
  clock_t tt;
  tt = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  tt = clock() - tt;
  double time_taken = ((double)tt) / CLOCKS_PER_SEC;
  return 0;
}