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 double pi = acos(-1.0);
double nGon(int N) {
  double proAngleVar;
  if (N % 4 == 0) {
    proAngleVar = pi * (180.0 / N) / 180;
  } else {
    proAngleVar = pi * (180.0 / (2 * N)) / 180;
  }
  double negX = 1.0e+99, posX = -1.0e+99, negY = 1.0e+99, posY = -1.0e+99;
  for (int j = 0; j < N; ++j) {
    double px = cos(2 * pi * j / N + proAngleVar);
    double py = sin(2 * pi * j / N + proAngleVar);
    negX = min(negX, px);
    posX = max(posX, px);
    negY = min(negY, py);
    posY = max(posY, py);
  }
  double opt2 = max(posX - negX, posY - negY);
  return (double)opt2 / sin(pi / N) / 2;
}
int main() {
  int N = 10;
  int t;
  cin >> t;
  while (t--) {
    cin >> N;
    cout << setprecision(8) << nGon(N * 2) << "\n";
  }
  return 0;
}