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;
inline long long int gcd(long long int a, long long int b) {
  long long int r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
inline long long int fpow(long long int n, long long int k,
                          int p = 1000000007) {
  long long int r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
inline long long int addmod(long long int a, long long int b) {
  return ((a % 1000000007) + (b % 1000000007)) % 1000000007;
}
inline long long int multmod(long long int a, long long int b) {
  return ((a % 1000000007) * (b % 1000000007)) % 1000000007;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    double y = 3.141592653589793238 / (2 * n), z = 1 / tan(y);
    printf("%0.8lf\n", z);
  }
  return 0;
}