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;

public class Simple_Polygon_Embedding {
    public static void main(String[] args) throws IOException {
        BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
        String test_numS=bf.readLine();
        int test_num=Integer.parseInt(test_numS);
        while(test_num>0){
            String T=bf.readLine();
            int t=2*Integer.parseInt(T);
            solve(t);
            test_num--;
        }
    }
    public static void solve(int t){
        int num=(t-4)/4;
        double up=0,down=0,zhijiao=Math.PI/2;
        double angle=((t-2)*Math.PI)/t;
        double len=1;
        if (num>0){
            len+=1*Math.sin(angle-zhijiao)*2;
            //System.out.println("+"+1*Math.sin(angle-zhijiao)+"len"+len);
            num--;
            down=zhijiao-(angle-zhijiao);
        }
        while(num>0){
            up=angle-zhijiao-down;
            len+=1*Math.sin(up)*2;
            down=zhijiao-up;
            num--;
        }
        System.out.println(len);
    }
}