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 JAVA):

Solution:

import java.util.Scanner;

public class third {
	static double polydiagonal(double n, double a) {

		// Side and side length cannot be negative
		if (a < 0 && n < 0) {
			return -1;
		}

		double rad = Math.toRadians(180 / n);
		return (double) (1 / Math.sin(rad));
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn = new Scanner(System.in);
		StringBuilder sb = new StringBuilder();
		int t = scn.nextInt();
		while (t-- >0) {
			int n = scn.nextInt();
			double x = Math.PI / (2 * n);
			double ans = Math.sin(x);
			ans = 1 / ans;
			double s = (Math.sqrt(ans * ans - 1.0));
			sb.append(s + "\n");
		}
		System.out.print(sb);

	}

}