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.util.Scanner;

public class cf1354_Div2_C2 {
			
	public static double[] rot(double[] s, double t) {
		return new double[]{s[0]*Math.cos(t) - s[1]*Math.sin(t), s[0]*Math.sin(t) + s[1]*Math.cos(t)};
	}
	
	public static double solve(double[][] s, double t, int n) {
		double min1 = Integer.MAX_VALUE; 
		double min2 = Integer.MAX_VALUE; 
		double max1 = Integer.MIN_VALUE;
		double max2 = Integer.MIN_VALUE;
				
		for (int i = 0; i < 2 * n; i++) {
			double[] c = rot(s[i], t);
			min1 = Math.min(c[0], min1);
			max1 = Math.max(c[0], max1);
			min2 = Math.min(c[1], min2);
			max2 = Math.max(c[1], max2);
		}
		
		//System.out.println(Math.min(Math.abs(min1 - max1), Math.abs(min2 - max2)));
		
		return Math.min(Math.abs(min1 - max1), Math.abs(min2 - max2));
		
	}
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);
		int tc = in.nextInt();
		double[] arr = new double[2];
		for ( ; tc > 0; tc--) {
			int n = in.nextInt();
			
			double[][] s = new double[2 * n][2];
						
			s[0][0] = 1 / (2 * Math.sin(Math.PI / (2 * n)));
			s[0][1] = 0;
			
			for (int i = 1; i < 2 * n; i++) {
				arr = rot(s[i - 1], Math.PI / n);
				s[i][0] = arr[0];
				s[i][1] = arr[1];
			}
					
			double l = 0;
			double r = Math.PI / 2 / n;
			while (r - l > (double)(1e-9)) {
				double mid1 = l + (r - l) / 3;
				double mid2 = r - (r - l) / 3;
				double s1 = solve(s, mid1, n);
				double s2 = solve(s, mid2, n);
				if (s1 < s2)
					l = mid1;
				else
					r = mid2;
			}
			
			System.out.println(solve(s, l, n));
		}
	}
}