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 long long N = 200005, INF = 2000000000000000000;
long long power(long long a, long long b, long long p) {
  if (a == 0) return 0;
  long long res = 1;
  a %= p;
  while (b > 0) {
    if (b & 1) res = (res * a) % p;
    b >>= 1;
    a = (a * a) % p;
  }
  return res;
}
vector<long long> prime;
bool is_composite[N];
void sieve(long long n) {
  memset(is_composite, false, sizeof(is_composite));
  for (long long i = 2; i < n; ++i) {
    if (!is_composite[i]) prime.push_back(i);
    for (long long j = 0; j < prime.size() && i * prime[j] < n; ++j) {
      is_composite[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
void print(bool n) {
  if (n) {
    cout << "YES";
  } else {
    cout << "NO";
  }
}
long double pi = 2 * acos(0.0);
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long T;
  cin >> T;
  while (T--) {
    long double n;
    cin >> n;
    n *= 2;
    long double ans = tan((pi * (n - 2)) / (2 * n));
    cout << setprecision(12);
    cout << ans << "\n";
  }
}