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 = 100005, 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;
}
bool prime[N] = {true};
long long sum[N] = {0};
void pre() {
  memset(prime, true, sizeof(prime));
  for (long long i = 2; i <= N; i++) {
    if (prime[i]) {
      for (long long j = i * i; j <= N; j += i) {
        prime[j] = false;
      }
    }
  }
  for (long long i = 2; i < N; i++) {
    if (prime[i]) sum[i] += i;
    sum[i + 1] = sum[i];
  }
  return;
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
int32_t main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  pre();
  long long t;
  cin >> t;
  while (t--) {
    long double n;
    cin >> n;
    double pi = 3.14159265358979323846;
    long double temp = 1 / tan(pi / (2 * n));
    cout << fixed << setprecision(8) << temp << "\n";
  }
}