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 mul(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
void input(long long a[], long long sz) {
  for (long long i = 0; i < sz; i++) cin >> a[i];
}
void print(long long a[], long long sz) {
  for (long long i = 0; i < sz; i++) {
    if (i == sz - 1)
      cout << a[i] << "\n";
    else
      cout << a[i] << " ";
  }
}
long long maxr(long long a[], long long sz) {
  long long ma;
  for (long long i = 0; i < sz; i++) {
    if (i == 0)
      ma = a[i];
    else if (a[i] > ma)
      ma = a[i];
  }
  return ma;
}
long long minr(long long a[], long long sz) {
  long long mi;
  for (long long i = 0; i < sz; i++) {
    if (i == 0)
      mi = a[i];
    else if (a[i] < mi)
      mi = a[i];
  }
  return mi;
}
long long isprm(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p = 1000000007) {
  return power(n, p - 2, p);
}
long long ncrMod(long long n, long long r, long long p = 1000000007) {
  if (r == 0) return 1;
  long long fac[n + 1];
  fac[0] = 1;
  for (long long i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % p;
  return (fac[n] * modInverse(fac[r], p) % p * modInverse(fac[n - r], p) % p) %
         p;
}
struct comp {
  bool operator()(const pair<long long, long long> &a,
                  const pair<long long, long long> &b) const {
    if (a.first == b.first) {
      return (a.second > b.second);
    }
    return (a.first < b.first);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    n = 2 * n;
    long double k = 3.1415926535 / n;
    long double ans = cos(k) / sin(k);
    cout << fixed << setprecision(15) << ans << "\n";
  }
  return 0;
}