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 N = 2e5 + 1;
long long int power(long long int x, long long int y, long long int p) {
  long long int temp;
  if (y <= 0) return 1;
  temp = power(x, y / 2, p);
  if (y % 2 == 0) {
    return (temp * temp) % p;
  } else {
    return (((x * temp) % p) * temp) % p;
  }
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int nCrModPFermat(long long int n, long long int r, long long int p) {
  if (r == 0) {
    return 1;
  }
  long long int fac[n + 1];
  fac[0] = 1;
  for (long long int i = 1; i <= n; i++) {
    fac[i] = fac[i - 1] * i % p;
  }
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long double k = (double)2 * n;
    long double temp = (double)360 / k;
    temp = (double)temp / 2;
    long double ans = tan(temp * 3.141592653 / 180);
    cout << fixed << setprecision(9) << 1 / ans << endl;
  }
  return 0;
}