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.InputStreamReader;
import java.util.ArrayList;
 

public class Main{
          
    static int n=0;
     
     public static void main(String args[]){
         BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
         
         try{ String line;
             while((line=br.readLine())!=null){if(line.isEmpty())break;
                String spl[] = line.split(",");
               int q = Integer.parseInt(spl[0]); if(q==-1)break;
               double x=(double)q/2;
               while(true){
                   x=x-(x*x*x-q)/(3*x*x);
                   if(Math.abs(x*x*x-q)<0.00001*q)break;
               }
                 System.out.println(x);
               
             } br.close();
            
         }catch(Exception e){e.printStackTrace();}         
     }  
    
     
     
}
 