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 <cstdio>
#include <cstdlib>
#include <cmath>
const static double EPS = 1e-5;

double cube( double x ) {
  return x * x * x;
}

int main() {

  while( 1 ) {
    int q;
    scanf( "%d", &q );
    if( q == -1 ) { break; }

    double x;
    for(
        x = q / 2.0;
        fabs( cube( x ) - q ) >= q * EPS;
        x = x - ( cube( x ) - q ) / ( 3 * x * x )
       );

    printf( "%.6f\n", x );

  }
  return EXIT_SUCCESS;
}