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;
long long M = 1e9 + 7;
long long powerm(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % M;
    y = y >> 1;
    x = (x * x) % M;
  }
  return res % M;
}
long long power(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long max3(long long a, long long b, long long c) {
  return max(max(a, b), c);
}
long long gcd(long long a, long long b) {
  if (b > a) return gcd(b, a);
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  double pi = acos(-1.0);
  cout << fixed << setprecision(10);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << 1.0 / tan(pi / (2.0 * n)) << '\n';
  }
  return 0;
}