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
#include<iostream>
#include <bits/stdc++.h>

using namespace std;

int main()
{
  double num, x0, x1 ;//?????°??£?¨?
  while(cin >> num)//??????????§?
  {
    if(num == -1) break;
    else{
      x0 = num/2.0 ;//?????????
      while(1){
        x1 = x0 - (pow(x0,3.0)-num)/(3*pow(x0,2.0)) ;//?????\????????????
        if(abs(pow(x1,3.0)-num)< 0.00001*num)//???????????¶?¢????
        {
          break ;
        }
        x0 = x1 ;//???????????????????????´????¬?????¨?????????????
      }
      cout<<fixed<<setprecision(8)<< x1 <<endl;
    }
  }
  return 0 ;
}