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.io.PrintWriter;
import java.util.StringTokenizer;


public class C1 {

    public static void main(String[] args) {
        new C1().run();
    }

    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    ;
    PrintWriter out = new PrintWriter(System.out);
    StringTokenizer tok = new StringTokenizer("");

    void run() {
        try {
            long timeStart = System.currentTimeMillis();
            solve();

            out.close();

            long timeEnd = System.currentTimeMillis();
            System.err.println("Time = " + (timeEnd - timeStart));
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    String readLine() throws IOException {
        return in.readLine();
    }

    String delimiter = " ";

    String readString() throws IOException {
        while (!tok.hasMoreTokens()) {
            String nextLine = readLine();
            if (null == nextLine) return null;

            tok = new StringTokenizer(nextLine);
        }

        return tok.nextToken(delimiter);
    }

    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }


    void solve() throws IOException {
        int t = readInt();
        int n;
        double ans;
        for (int i = 0; i < t; i++){
            n = readInt();
            //if (Math.pow(2, Math.log10(n) / Math.log10(2)) == n){
            ans = 1 / Math.tan(Math.PI / (2*n));
            //}
            //else
              //  ans = Math.pow(Math.sin(Math.PI / (2*n)), -1);
            out.println(ans);
        }
    }


}