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<bits/stdc++.h>
#include<chrono>

using namespace std;
using namespace std::chrono;

/// ----------------------------- (Debug) ------------------------------
#define sim template < class c
#define ris return * this
#define dor > debug & operator <<
#define eni(x) sim > typename enable_if<sizeof dud<c>(0) x 1, debug&>::type operator<<(c i) {
sim > struct rge { c b, e; }; sim > rge<c> range(c i, c j) { return rge<c>{i, j}; }
sim > auto dud(c* x) -> decltype(cerr << *x, 0); sim > char dud(...);
struct debug {
#ifndef ONLINE_JUDGE
eni(!=) cerr << boolalpha << i; ris; }
eni(==) ris << range(begin(i), end(i));}
sim, class b dor(pair < b, c > d) {ris << "(" << d.first << ", " << d.second << ")";}
sim dor(rge<c> d) {*this << "["; for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it; ris << "]";}
#else
sim dor(const c&) { ris; }
#endif
};
vector<char*> tokenizer(const char* args) {
    char *token = new char[111]; strcpy(token, args); token = strtok(token, ", ");
    vector<char*> v({token});
    while(token = strtok(NULL,", ")) v.push_back(token);
    return reverse(v.begin(), v.end()), v;
}
void debugg(vector<char*> args) { cerr << "\b\b "; }
template <typename Head, typename... Tail>
void debugg(vector<char*> args, Head H, Tail... T) {
    debug() << " [" << args.back() << ": " << H << "] ";
    args.pop_back(); debugg(args, T...);
}
#define harg(...) #__VA_ARGS__
#ifndef ONLINE_JUDGE
#define dbg(...) { debugg(tokenizer(harg(__VA_ARGS__, \b\t-->Line)), __VA_ARGS__, __LINE__); cerr << endl;}
#else
#define dbg(...) { }
#endif
/// -----------------------------------------------------------------------


typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;
typedef vector<pi> vpi; 
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)    
#define F first
#define S second
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define print(x) cout << x << "\n";
#define REP(i,a,b) for(i=a;i<=b;i++)
#define RAP(i,a,b) for(i=a;i>=b;i--)
#define spa <<" "<<
#define all(x) (x).begin(), (x).end()
int mod=1000000007;
const int MX=0x3f3f3f3f;

int main(){
    IOS;
    ll x,i,j;
    cin>>x;
    if(x==1) {
        cout<<0 spa -1;
        return 0;
    }
    ll p[131];
    p[0]=0;
    REP(i,1,130) p[i]=pow(i,5);
    i=1;
    j=2;
    while(j<131){
        if(i==j||p[i]+p[j]<x) j++;
        else if(p[i]+p[j]>x) i++;
        else {
            cout<<j<<" -"<<i;
            return 0;
        }
    }
    i=1;
    j=2;
    while(j<131){
        if(p[j]-p[i]>x) i++;
        else if(p[j]-p[i]<x) j++;
        else{
            cout<<j<<" "<<i;
            return 0;
        }
    }
    //cout<<1 spa 3;
    return 0;
}