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 = 1e5 + 5;
const long long mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int dir[4][2] = {-1, 0, 1, 0, 0, -1, 0, 1};
const int exdir[4][2] = {1, 1, 1, -1, -1, 1, -1, -1};
long long qpow(long long x, long long y) {
  long long ans = 1, t = x;
  while (y > 0) {
    if (y & 1) ans *= t, ans %= mod;
    t *= t, t %= mod;
    y >>= 1;
  }
  return ans % mod;
}
double pi = acos(-1.0);
void solve() {
  int n;
  cin >> n;
  printf("%.8f", 1 / tan(pi / (2 * n)));
}
int main() {
  int t;
  cin >> t;
  while (t--) solve(), cout << '\n';
  return 0;
}