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;
vector<int> a(200005);
int n;
int k;
bool calc(int mx) {
  int ok = k;
  if (ok % 2 == 0) {
    k = k / 2;
    int ct = 0;
    int ls = -3;
    for (int i = 0; i < n; i++) {
      if (i - ls >= 2 && a[i] <= mx) {
        ls = i;
        ct++;
        if (i < n - 1 && ct >= k) {
          k = ok;
          return true;
        }
      }
    }
    ct = 0;
    ls = -3;
    for (int i = 1; i < n; i++) {
      if (i - ls >= 2 && a[i] <= mx) {
        ls = i;
        ct++;
        if (ct >= k) {
          k = ok;
          return true;
        }
      }
    }
    k = ok;
    return false;
  } else {
    k = k / 2;
    int ct = 0;
    int ls = -3;
    for (int i = 1; i < n; i++) {
      if (i - ls >= 2 && a[i] <= mx) {
        ls = i;
        ct++;
        if (i < n - 1 && ct >= k) {
          k = ok;
          return true;
        }
      }
    }
    k++;
    ct = 0;
    ls = -3;
    for (int i = 0; i < n; i++) {
      if (i - ls >= 2 && a[i] <= mx) {
        ls = i;
        ct++;
        if (ct >= k) {
          k = ok;
          return true;
        }
      }
    }
    k = ok;
    return false;
  }
}
void solve() {
  double n;
  cin >> n;
  double pi = 3.1415926535897932384626;
  cout.precision(10);
  cout << cos(pi / (4 * n)) / sin(pi / (2 * n)) << endl;
}
void querySolve() {
  int n;
  cin >> n;
  while (n--) {
    solve();
  }
}
int main() { querySolve(); }