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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.*;


public class main {
    
         
    static FastReader in;
    static StringBuilder out;
    public  static void main(String[] args) {
        out = new StringBuilder();
        in = new FastReader();
        int t=in.nextInt();

        while(t-->0){
            int n=in.nextInt();
            double extAngle=Math.PI/n;
            double ans=0;
            for(int i=1;i<=n;i++){
                ans+=Math.sin(extAngle*i);
            }
            out.append(ans).append('\n');
        }        
        System.out.println(out);
    }
    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;
        }
    }
 
    static long gcd(long a, long b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }
 
    static long findGCD(int arr[]) {
        long result = arr[1];
        for (int i = 1; i < arr.length - 1; i++)
            result = gcd(arr[i], result);
 
        return result;
    }
 
    private static long[] getLongArrar(int size) {
        long arr[] = new long[size + 2];
        for (int i = 0; i <= size; i++) {
            arr[i] = in.nextLong();
        }
        return arr;
    }
 
    private static String[] getStringArray(int size) {
        String arr[] = new String[size + 2];
        for (int i = 1; i <= size; i++) {
            arr[i] = in.next();
        }
        return arr;
    }
 
    private static int[] getIntArray(int size) {
        int arr[] = new int[size + 2];
        for (int i = 1; i <= size; i++) {
            arr[i] = in.nextInt();
        }
        return arr;
    }
 
    private static long[] CumulativeOfArrayCreator(int arr[], int size) {
        long cumulative[] = new long[size];
        cumulative[0] = arr[0];
        for (int i = 1; i < size; i++) {
            cumulative[i] = cumulative[i - 1] + arr[i];
        }
        return cumulative;
    }
 
    
}