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;
int n;
int read() {
  int s = 0, w = 1;
  char c = getchar();
  while ((c < '0' || c > '9') && c != '-') c = getchar();
  if (c == '-') w = -1, c = getchar();
  while (c <= '9' && c >= '0') s = (s << 3) + (s << 1) + c - '0', c = getchar();
  return s * w;
}
double ans = 0.0, pi = acos(-1);
void work() {
  n = read();
  ans = 0.0;
  double t = pi / 4;
  while (-(pi / 2 + 1e-7) <= t) t -= pi / n;
  t += pi / n;
  while (pi / 2 + 1e-7 >= t) ans += cos(t), t += pi / n;
  printf("%.10lf\n", ans);
}
signed main() {
  int T = read();
  while (T--) work();
  return 0;
}