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.math.BigDecimal;
import java.math.BigInteger;
import java.math.RoundingMode;
import java.time.chrono.IsoChronology;
import java.util.*;

public class div {
    private static Scanner sc = new Scanner(System.in);
    private static Map<Integer, Integer> m = new TreeMap<>();
    private static List<Character> l = new ArrayList<>();

    private static int sum = 0;

    public static String sortString(String inputString) {
        // convert input string to char array
        char tempArray[] = inputString.toCharArray();

        // sort tempArray
        Arrays.sort(tempArray);

        // return new sorted string
        return new String(tempArray);
    }

    public static void main(String[] args) {
        int t = sc.nextInt();
        sc.nextLine();

        for (int i = 0; i < t; i++) {
            int n = sc.nextInt();
            double radius = 1 / (2 * Math.sin((Math.PI / (2 * n))));
            double d = 2 * Math.sqrt(((radius * radius - 0.5 * 0.5)));

            BigDecimal bd = new BigDecimal(Double.toString(d));
            bd = bd.setScale(9, RoundingMode.HALF_UP);
            System.out.println(bd.doubleValue());
            
        }

    }
}