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.*;
import java.util.*;

public class Q1 {

    public static void main(String[] args) {

        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
     //   System.out.println(Math.tan(45));
        int t =in.nextInt();
        while(t-->0){
          double n=in.nextInt();
          double val=(90*1.0/n)*(Math.PI/180);
           //System.out.println(val);
          double angle = Math.tan(val);
        //   System.out.println(angle);
          double ans =0.5/angle;
        //    System.out.println(ans);
          out.println(ans+ans);
        }
        out.close();
    }


    static long gcd(long a, long b) {
        return b == 0 ? a : gcd(b, a % b);
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        static int[][] packU(int n, int[] from, int[] to) {
            int[][] g = new int[n][];
            int[] p = new int[n];
            for (int f : from)
                p[f]++;
            for (int t : to)
                p[t]++;
            for (int i = 0; i < n; i++)
                g[i] = new int[p[i]];
            for (int i = 0; i < from.length; i++) {
                g[from[i]][--p[from[i]]] = to[i];
                g[to[i]][--p[to[i]]] = from[i];
            }
            return g;
        }

        public int[] shuffle(int[] arr) {
            Random r = new Random();
            for (int i = 1, j; i < arr.length; i++) {
                j = r.nextInt(i);
                arr[i] = arr[i] ^ arr[j];
                arr[j] = arr[i] ^ arr[j];
                arr[i] = arr[i] ^ arr[j];
            }
            return arr;
        }

        public InputReader() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public char nextChar() {
            return next().charAt(0);
        }

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

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

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


    }

}