Problem: Takahashi has a water bottle with the shape of a rectangular prism whose base is a square of side a~\mathrm{cm} and whose height is b~\mathrm{cm}. (The thickness of the bottle can be ignored.)

We will pour x~\mathrm{cm}^3 of water into the bottle, and gradually tilt the bottle around one of the sides of the base.

When will the water be spilled? More formally, find the maximum angle in which we can tilt the bottle without spilling any water.

Constraints

* All values in input are integers.
* 1 \leq a \leq 100
* 1 \leq b \leq 100
* 1 \leq x \leq a^2b

Input

Input is given from Standard Input in the following format:


a b x


Output

Print the maximum angle in which we can tilt the bottle without spilling any water, in degrees. Your output will be judged as correct when the absolute or relative error from the judge's output is at most 10^{-6}.

Examples

Input

2 2 4


Output

45.0000000000


Input

12 21 10


Output

89.7834636934


Input

3 1 8


Output

4.2363947991

Solution (in JAVA):

Solution:
import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        // Your code here!
        
        Scanner sc = new Scanner(System.in);
        
        int a = sc.nextInt();
        int b = sc.nextInt(); 
        int x = sc.nextInt(); 
        
        WaterBottle wb = new WaterBottle(a, b, x);
        
        System.out.println(wb.search());
        
    }
    
}

class WaterBottle {
    private int a;
    private int b;
    private int x;
    
    public WaterBottle(int a, int b, int x){
        this.a = a;
        this.b = b;
        this.x = x;
    }
    
    public boolean judge(double Theta){
        
        double cos = Math.cos(Math.toRadians(Theta));
        double sin = Math.sin(Math.toRadians(Theta));
        
        if( (a * cos / sin) <= (double)b){
            return (  a * a * b - cos * a * a * a / sin / 2  >= (double)x);
        } else{
            return (a * b * b * sin / cos / 2 >= (double)x);
        }
        
    }
    
	public double search() {
		double left = 0;
		double right = 90;
		for(int i = 0; i <= 100; i++){
			double mid = (left + right) / 2;
			
			if(judge(mid)){
			    // 溢れなかったのでより小さい角度
			    right = mid;
			}else{
			    // 溢れたのでより大きい角度
			    left = mid;
			}
		}
		return 90.0 - right;
	}
    
}