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 C1 {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		double arr[] = new double[t];
		for (int i = 0; i < t; i++) {
			int n = sc.nextInt();
			double angle = (2*n - 2) * Math.PI/(2*n);
			double len = 1;
//			System.out.println(angle);
			double a = 0;
			for (int j = 0; j < (n - 2)/2; j++) {
				a += Math.PI - angle;
//				System.out.println(a);
				len += 2 * Math.cos(a);
			}
			arr[i] = len;
		}
		for (int i = 0; i < t; i++) {
			System.out.println(arr[i]);
		}
	}

}