The solution of $ x ^ 3 = q $ is to calculate the recurrence formula $ x_ {n + 1} = x_n-\ frac {x_ {n} ^ 3 --q} {3x_ {n} ^ 2} $ It can be calculated approximately with.

Put a positive number $ \ frac {q} {2} $ in $ x_1 $

$ x_2 = x_1-\ frac {x_ {1} ^ 3 --q} {3x_ {1} ^ 2} $, $ x_3 = x_2-\ frac {x_ {2} ^ 3 --q} {3x_ {2} ^ Calculate as 2} $,….

While doing this calculation

When the value of $ | x ^ 3 --q | $ becomes small enough, stop the calculation and use the last calculated $ x_n $ as an approximate solution of $ x ^ 3 = q $.

Follow this method to create a program that outputs an approximation of the cube root of $ q $ for the input positive integer $ q $. However, use $ | x ^ 3-q | <0.00001 q $ to determine that it is "small enough".



input

Multiple datasets are given. For each dataset, $ q $ ($ 1 \ leq q <2 ^ {31} $) (integer) is given on one line. The end of the input is -1.

The number of datasets does not exceed 50.

output

Print $ x $ (real number) on one line for each dataset. The output result may contain an error of 0.00001 or less.

Example

Input

15
15
-1


Output

2.466212
2.466212
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

class Main{
	public static void main(String[] args) {
		BufferedReader bfr=new BufferedReader(new InputStreamReader(System.in));
		String str="";
		int input=0;
		double ans=0;
		try {
			while(true) {
				str=bfr.readLine();
				input=Integer.parseInt(str);
				if(input<0) {
					System.exit(0);
				}
				ans=(double)input;
				ans=kotae(ans/2,input);
				System.out.println(ans);
			}
		}catch(IOException e) {
		}
	}

	public static double kotae(double a,int b) {		//a 漸化式のこう　b3乗したあたい
		double newans=0;
		double bb=(double) b;

		newans=a-(a*a*a-bb)/(3*a*a);

		if(Math.abs(newans*newans*newans-bb)<0.00001*bb) {
			return newans;
		}
		else {
			newans=kotae(newans,b);
		}

		return newans;
	}
}
