#include <iostream>
using namespace std;

int main(){

    double x;

    cin >> x;

    //二分法
    double mid, l = -10000, r= 10000;
    while(r-l>1e-8){  //经验判断，比要保留的有效位数多两个精度
        mid = (l+r)/2;
        if(mid * mid * mid >=x ) r = mid;
        else l =  mid;
    }

    printf("%f\n",mid);
    return 0;
}