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.*;

import java.io.*;

public class Main {
	static final int M = 1000000007;
	static FastReader in = new FastReader();
//	static PrintWriter out = new PrintWriter(System.out);
//	static Scanner in = new Scanner(System.in);
//	File file = new File("input.txt");
//	Scanner in = new Scanner(file);
//	PrintWriter out = new PrintWriter(new FileWriter("output.txt"));
//	Arrays.sort(arr, Comparator.comparingDouble(o -> o[0]));
	
	public static void main(String[] args) {
		
		int t = in.nextInt();
		while(t-- > 0) solve();
//		 solve();
		
	}
	
	static void solve() {
		int n = in.nextInt();
		System.out.println(1/Math.tan(Math.PI/(double)(2*n)));
	}
	
	static void sieve(int n, boolean[] prime)
    {
        for (int i = 0; i <= n; i++)
            prime[i] = true;
 
        for (int p = 2; p * p <= n; p++) 
        {
            if (prime[p] == true) 
            {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
    }
	static int gcd(int a, int b) {
		if(b==0) return a;
		return gcd(b, a%b);
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new
					InputStreamReader(System.in));
		}

		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}
}