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;
int bpow(int a, int b) {
  long double res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a);
    a = (a * a);
    b >>= 1;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    if (n == 2) {
      cout << fixed << setprecision(9) << 1 << "\n";
      continue;
    }
    if (n % 2 == 0) {
      long double y = 2 * n;
      long double a = ((y - 2) * 90 * 1.0) / y;
      long double b = 90.0 - a;
      a = a * 3.1415926535897932385 / 180.0;
      b = b * 3.1415926535897932385 / 180.0;
      long double c = (sin(a) * 0.5) / sin(b);
      cout << fixed << setprecision(9) << 2 * c << "\n";
    } else {
      long double y = 2 * n;
      long double a = ((y - 2) * 90 * 1.0) / y;
      long double b = 180.0 - 2 * a;
      a = a * 3.1415926535897932385 / 180.0;
      b = b * 3.1415926535897932385 / 180.0;
      long double c = (sin(a) * 1.0) / sin(b);
      cout << fixed << setprecision(9) << 2 * c << "\n";
    }
  }
  return 0;
}