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.*;
import java.lang.*;
public class Codechef {
	PrintWriter out;
    StringTokenizer st;
    BufferedReader br;
 	
 	class Pair implements Comparable<Pair>
	{
	    int f;
	    int s;
 
	    Pair(int t, int r) {
	       f = t;
	       s = r;
	    }
	    public int compareTo(Pair p)
	    {
	    	if(this.f!=p.f)
	    		return this.f-p.f;
 
	    	return this.s-p.s;
	    }
 	}
 // 	class Sort implements Comparator<String> 
	// { 
	//     public int compare(String a, String b) 
	//     { 
	//     	return (a+b).compareTo(b+a);
	//     } 
	// }
 
	String ns() {
        while (st == null || !st.hasMoreElements()) {
 
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
 
    String nextLine() throws Exception {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            throw new Exception(e.toString());
        }
        return str;
    }
 
    int nextInt() {
        return Integer.parseInt(ns());
    }
 
    long nextLong() {
        return Long.parseLong(ns());
    }
 
    double nextDouble() {
        return Double.parseDouble(ns());
    }
	
	int upperBound(ArrayList<Integer> a,int key)
	{
		int l=0,r=a.size()-1;
		int i=-1;
 
		while(l<=r)
		{
			int mid=(l+r)/2;
 
			if(a.get(mid)<=key)	
				l=mid+1;
			
 
			else{
				i=mid;
				r=mid-1;
			}
		}
		if(i!=-1)
			return a.get(i);
		return i;
	}
 
	long power(long x,long y)
	{
	    long ans=1;
	    
	    while(y!=0)
	    {
	        if(y%2==1) ans=(ans*x)%mod;
	        x=(x*x)%mod;
	        y/=2;
	    }
	    return ans%mod;
	}
 
	long mod= 1000000007;
	long gcd(long x ,long y)
	{
		if(y==0)
			return x;
		return gcd(y,x%y);
	}

	// ArrayList a[];
	// int vis[];
	// ArrayList<Integer> list;
	// int dfs(int ver)
	// {
	// 	ArrayList<Integer> l=a[ver];
	// 	int cnt=0;
	// 	for(int v:l)
	// 	{
	// 		if(vis[v]==0){
	// 			vis[v]=vis[ver]+1;
	// 			cnt+=dfs(v)+1;
	// 		}
	// 	}
	// 	list.add(vis[ver]-cnt-1);
	// 	return cnt;
	// }

	void solve() throws IOException{
		int t=nextInt();
		while(t-->0){
			int n=nextInt();
			out.println(1/Math.tan(Math.PI/(2*n)));
			//out.println(1/Math.sin((Math.PI/(2*n))));
		}

	}
	
    void run() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        long s = System.currentTimeMillis();
		solve();
		out.flush();
 
		out.close();
	}
 
	public static void main(String args[]) throws IOException {
		new Codechef().run();
	}
  
}