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 read() {
  int x = 0, flag = 0;
  char ch = getchar();
  if (ch == '-') flag = 1;
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x *= 10, x += ch - '0', ch = getchar();
  if (flag) return -x;
  return x;
}
int main() {
  int test;
  cin >> test;
  while (test--) {
    double i;
    double sum = 0;
    int n;
    cin >> n;
    double t = tan(acos(-1) / (2 * n));
    printf("%.9lf\n", 1 / t);
  }
}