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 MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI = acos(-1);
const int INF = 1e9;
int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
long long gcd(long long a, long long b) {
  if (a == 0) return (b);
  return (gcd(b % a, a));
}
long long lcm(long long a, long long b) { return (a / gcd(a, b) * b); }
bool isprime(long long x) {
  if (x == 1) return (false);
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return (false);
  }
  return (true);
}
long long dp[1000000];
int div(long long i) {
  long long count = 0;
  for (long long j = 1; j * j <= i; j++) {
    if (i % j == 0) {
      count++;
      if (i / j != j) count++;
    }
  }
  return (count);
}
void solve() {
  long long n;
  cin >> n;
  double ans = 1 / tan(PI / (2 * n));
  cout << fixed << setprecision(16) << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return (0);
}