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 = 2e5 + 20;
const int MAX = 10000007;
inline long long read() {
  char c = getchar();
  long long x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0', c = getchar();
  }
  return x * f;
}
inline void out(int x) {
  if (x > 9) out(x / 10);
  putchar(x % 10 + '0');
}
int q[N];
char s[N];
const double P = acos(-1);
int main() {
  int t;
  t = read();
  while (t--) {
    double n;
    cin >> n;
    double jo = 2 * P / (n) / 2;
    double x = 0.5 / sin(jo / 2);
    double jo1 = round(n / 4) * jo;
    double jo2 = P / 4.0;
    double jo3 = P / 2 - jo1;
    double ans1 = sin(jo1) * x / sin(jo2);
    double ans2 = sin(jo3) * x / sin(jo2);
    printf("%.8f\n", ans1 + ans2);
  }
  return 0;
}