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 <bits/stdc++.h>
using namespace std;
double temp(double,double);

double temp(double x,double q){
    x-=(x*x*x-q)/(3*x*x);
    return x;
}

int main(){
    double x[50],q;
    int i;
    for(i=0;i<50;i++){
        cin>>q;
        if(q==-1)break;
        x[i]=q/2;
        do{
            x[i]=temp(x[i],q);
        }while(fabs(x[i]*x[i]*x[i]-q)>=1e-5*q);
    }
    for(int j=0;j<i;j++){
        printf("%.6f\n",x[j]);
    }
}