#include <bits/stdc++.h>

using namespace std;

int main()
{
    int c;
    cin>>c;
    /// rezolvam ec. (n*n*n + n)/2 = c, folosind ec de gr 3 redusa
    /// cum avem 2 sol complexe, calculam doar cea reala
    double v1 = cbrt(c*1.0+sqrt(1.0/27.0+c*c*1.0));
    double v2 = cbrt(c*1.0-sqrt(1.0/27.0+c*c*1.0));
    double n = v1+v2;
    /// in mod mormal, avem n o valoare intreaga, dar datorita
    /// impreciziei tipului double/float, rotunjim si verificam
    int rez = 0;
    int eroare = 1;
    for(int i=n-eroare;i<=n+eroare;++i)
        if((i*i*i+i)/2 == c){
            rez = i;
            break;
        }
    cout<<rez;
    return 0;
}