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.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		// double q;
		while (true) {
			double q = in.nextDouble();
			if (q == -1)
				break;
			double x = q / 2;
			while (Math.abs(x * x * x - q) > 0.00001 * q)
				x = Math.abs(x - ((x * x * x - q) / (3 * x * x)));
			System.out.printf("%.6f\n", x);
		}
		// while ((q = in.nextInt()) >= 1) {
		// double xn = q / 2.0;
		// double checkVal = xn * xn * xn - q;
		// while (Math.abs(checkVal) >= 0.00001 * q) {
		// checkVal = xn * xn * xn - q;
		// xn = xn - (checkVal) / (xn * xn * 3.0);
		// }
		// System.out.printf("%f", xn);
		// System.out.println();
		//
		// }
	}
}