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:
// Java program to find the smallest window containing 
// all characters of a pattern. 
import java.io.*;
import java.util.*;
public class Solution { 

    static final int MAX_CHARS = 256; 

    // Function to find smallest window containing 
    // all distinct characters 
    static String findSubString(String str) 
    { 
        int n = str.length(); 

        // Count all distinct characters. 
        int dist_count = 0; 

        boolean[] visited = new boolean[MAX_CHARS]; 
        Arrays.fill(visited, false); 
        for (int i = 0; i < n; i++) { 
            if (visited[str.charAt(i)] == false) { 
                visited[str.charAt(i)] = true; 
                dist_count++; 
            } 
        } 

        // Now follow the algorithm discussed in below 
        // post. We basically maintain a window of characters 
        // that contains all characters of given string. 
        int start = 0, start_index = -1; 
        int min_len = Integer.MAX_VALUE; 

        int count = 0; 
        int[] curr_count = new int[MAX_CHARS]; 
        for (int j = 0; j < n; j++) { 
            // Count occurrence of characters of string 
            curr_count[str.charAt(j)]++; 

            // If any distinct character matched, 
            // then increment count 
            if (curr_count[str.charAt(j)] == 1) 
                count++; 

            // if all the characters are matched 
            if (count == dist_count) { 
                // Try to minimize the window i.e., check if 
                // any character is occurring more no. of times 
                // than its occurrence in pattern, if yes 
                // then remove it from starting and also remove 
                // the useless characters. 
                while (curr_count[str.charAt(start)] > 1) { 
                    if (curr_count[str.charAt(start)] > 1) 
                        curr_count[str.charAt(start)]--; 
                    start++; 
                } 

                // Update window size 
                int len_window = j - start + 1; 
                if (min_len > len_window) { 
                    min_len = len_window; 
                    start_index = start; 
                } 
            } 
        } 
        // Return substring starting from start_index 
        // and length min_len 
        return str.substring(start_index, start_index + min_len); 
    } 

    // Driver code 
    public static void main(String args[])throws IOException 
    { 
        BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
        int t=Integer.parseInt(in.readLine());
        StringBuilder ans=new StringBuilder();
        while(t-->0){
            int n=Integer.parseInt(in.readLine());
            if(n==2){
                ans.append(1+"\n");
                continue;
            }
            n*=2;
            double get=Math.tan(Math.toRadians(90)-Math.toRadians((double)180.0/(double)n));
            ans.append(get+"\n");
        }
        System.out.println(ans);
    } 
} 
// This code is contributed by Sumit Ghosh