Problem:
Score : 400 points

Problem Statement

Give a pair of integers (A, B) such that A^5-B^5 = X.
It is guaranteed that there exists such a pair for the given integer X.

Constraints

1 \leq X \leq 10^9

X is an integer.

There exists a pair of integers (A, B) satisfying the condition in Problem Statement.

Input

Input is given from Standard Input in the following format:

X

Output

Print A and B, with space in between.
If there are multiple pairs of integers (A, B) satisfying the condition, you may print any of them.

A B

Sample Input 1

33

Sample Output 1

2 -1

For A=2 and B=-1, A^5-B^5 = 33.

Sample Input 2

1

Sample Output 2

0 -1

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <iomanip>
#include <utility>
#include <tuple>
#include <functional>
#include <bitset>
#include <cassert>
#include <complex>
#include <time.h>
#define ll long long
#define double long double
#define itn int
#define endl '\n'
#define co(ans) cout<<ans<<endl;
#define COYE cout<<"YES"<<endl;
#define COYe cout<<"Yes"<<endl;
#define COye cout<<"yes"<<endl;
#define CONO cout<<"NO"<<endl;
#define CONo cout<<"No"<<endl;
#define COno cout<<"no"<<endl;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define RREP(i,n)  FFOR(i,1,n)
#define ALL(V) ((V).begin(),(V).end())
#define SORT(V) sort((V).begin(),(V).end())
#define REVERSE(V) reverse((V).begin(),(V).end())
#define INF ((1LaL<<62)-(1LL<<31))
#define EPS 1e-10
#define PI 3.141592653589793238
#define MOD 1000000007
#define MAX 5100000
template<class T>bool chmax(T &a,const T &b){if(a<b){a=b;return 1;}return 0;}
template<class T>bool chmin(T &a,const T &b){if(b<a){a=b;return 1;}return 0;}
using namespace std;

int main(){
  unsigned long long X;
  cin>>X;
  
  unsigned long long A[11000];
  REP(i,1000){
    A[i]=i*i*i*i*i;
  }
  
  for(int i=0; i<1000; ++i){
    for(int j=0; j<=i; ++j){
      if(A[i]-A[j]==X){
        cout<<i<<" "<<j<<endl;
        return 0;
      }
      else if(A[i]+A[j]==X){
        j*=-1;
        cout<<i<<" "<<j<<endl;
        return 0;
      }
    }
  }
  
  return 0;
}