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;
inline int read() {
  char ch = getchar();
  int nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  int ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
inline int min(int x, int y, int z) { return min(x, min(y, z)); }
inline int max(int x, int y, int z) { return max(x, max(y, z)); }
inline int add(int x, int y) {
  return x + y >= 998244353 ? x + y - 998244353 : x + y;
}
inline int add(int x, int y, int z) { return add(add(x, y), z); }
inline int sub(int x, int y) { return x - y < 0 ? x - y + 998244353 : x - y; }
inline int mul(int x, int y) { return 1LL * x * y % 998244353; }
inline int mul(int x, int y, int z) { return mul(mul(x, y), z); }
void work() {
  int n = read();
  double a = acos(-1) / 2.0 / n;
  double x = 1 / sin(a);
  a = (acos(-1) / 2 - acos(-1) / n * (n / 2)) / 2;
  double ans = x * cos(a);
  printf("%.6lf\n", ans);
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}