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) {
		Scanner sc = new Scanner(System.in);
		int a =sc.nextInt();
		int b =sc.nextInt();
		int x =sc.nextInt();
		
		//溢れる瞬間には、水は底辺b,高さ1以下の直角三角形を底面とする高さaの三角柱
		//この体積が、a*a*b
		
		double ans=0;
		double tmp=0;
		
		//体積が、a*a*b/2いかのとき、
		if(x<=(a*a*b/2)) {
			tmp=((double)2*x)/(a*b*b);
			ans = 90.0 - Math.atan(tmp)*180.0/Math.PI;//度数に直す
			
		}else {
			//倒した時に三角形ができない。台形になる場合
			tmp=((double)2*(a*a*b-x))/(a*a*a);
			ans = Math.atan(tmp)*180.0/Math.PI;//度数に直す
		}
		
		System.out.println(ans);
	}
}