Problem: Fifa and Fafa are sharing a flat. Fifa loves video games and wants to download a new soccer game. Unfortunately, Fafa heavily uses the internet which consumes the quota. Fifa can access the internet through his Wi-Fi access point. This access point can be accessed within a range of r meters (this range can be chosen by Fifa) from its position. Fifa must put the access point inside the flat which has a circular shape of radius R. Fifa wants to minimize the area that is not covered by the access point inside the flat without letting Fafa or anyone outside the flat to get access to the internet.

The world is represented as an infinite 2D plane. The flat is centered at (x1, y1) and has radius R and Fafa's laptop is located at (x2, y2), not necessarily inside the flat. Find the position and the radius chosen by Fifa for his access point which minimizes the uncovered area.

Input

The single line of the input contains 5 space-separated integers R, x1, y1, x2, y2 (1 ≤ R ≤ 105, |x1|, |y1|, |x2|, |y2| ≤ 105).

Output

Print three space-separated numbers xap, yap, r where (xap, yap) is the position which Fifa chose for the access point and r is the radius of its range. 

Your answer will be considered correct if the radius does not differ from optimal more than 10 - 6 absolutely or relatively, and also the radius you printed can be changed by no more than 10 - 6 (absolutely or relatively) in such a way that all points outside the flat and Fafa's laptop position are outside circle of the access point range.

Examples

Input

5 3 3 1 1


Output

3.7677669529663684 3.7677669529663684 3.914213562373095


Input

10 5 5 5 15


Output

5.0 5.0 10.0

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.lang.*;

public class File {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
		
		// Flat radius:
		double R = sc.nextDouble(); 
		// Flat coordinates:
		double x1 = sc.nextDouble();
		double y1 = sc.nextDouble();
		
		// Fafa coordinates:
		double x2 = sc.nextDouble();
		double y2 = sc.nextDouble();
		
		if (!contains(x1, y1, R, x2, y2)) {
			System.out.println(x1 + " " + y1 + " " + R);
			return;
		}
		
		// If Fafa is in the center of the flat.
		if (x1 == x2 && y1 == y2) {
			double dR = R / 2.0;
			double dx = x1;
			double dy = y2 - dR;
			
			System.out.println(dx + " " + dy + " " + dR);
			return;
		}
		
		// Get slope of x2, y2 from origin.
		// Be careful of 0's.
		
		// Slope = x2 / y2
		double distance = getLength(x1, y1, x2, y2);
		
		double diameter = distance + R;
		double answerR = diameter / 2.0;
		
		// Similar triangles.
		/*
		x2 - answerX 
		/
		x2 - x1
		
		equals to 
				
		answerR
		/
		distance
		
		AND,
		
		y2 - answerY 
		/ 
		y2 - y1 
		
		equals to 
		
		answerR
		/ 
		distance
		
		=>
		
		(x1 - answerX) = (x2 - x1) * (answerR) / distance
		
		answerX = x1 - (x2 - x1) * (answerR) / distance
		
		answerY = y1 - (y2 - y1) * (answerR) / distance
		*/
		
		double answerX = x2 - (x2 - x1) * answerR / distance;
		double answerY = y2 - (y2 - y1) * answerR / distance;
		
		System.out.println(answerX + " " + answerY + " " + answerR);
	}
	
	public static double getLength(double x1, double y1, double x2, double y2) {
		double d_x = Math.abs(x1 - x2);
		double d_y = Math.abs(y1 - y2);
		
		return Math.sqrt(d_x*d_x + d_y*d_y);
	}
	
	public static boolean contains(double x1, double y1, double R, double x2, double y2) {
		double d_x = Math.abs(x2 - x1);
		double d_y = Math.abs(y2 - y1);
		double distance = Math.sqrt(d_x*d_x + d_y*d_y);
		
		return distance < R;
	}
}