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 mx = 1000011;
const long long mod = 1e9 + 7;
string s;
int n, q;
int tree[mx];
void update(int x, int val) {
  while (x <= n) {
    tree[x] += val;
    x += (x & -x);
  }
}
int query(int x) {
  int sum = 0;
  while (x > 0) {
    sum += tree[x];
    x -= (x & -x);
  }
  return sum;
}
void solve() {
  int n;
  cin >> n;
  if (n == 2) {
    cout << 1 << endl;
    return;
  }
  n = 2 * n;
  double angle = (n - 2) * 1.0 * 180.0;
  angle = angle / (n * 1.0);
  angle -= 90.0;
  double pi = acos(-1.0) / 180.0;
  double last = 0.0;
  double ans = 0;
  for (int i = 0; i < (n - 4) / 4; i++) {
    double angles = angle - last;
    ans += 2.0 * sin(angles * pi);
    last = 90.0 - (angle - last);
  }
  ans += 1.0;
  cout << setprecision(14) << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}