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 <string>
#include <sstream>
#include <utility>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <cmath>
#include <functional>
#include <chrono>
#include <cstring>
#include <limits.h>

using namespace std;
typedef long long lint;
typedef vector<int> vi;
typedef pair<int,int> pii;
typedef vector<lint> vl;
typedef pair<lint,lint> pll;

#define mpr make_pair


int main()
{
    ios_base::sync_with_stdio(0);    cin.tie(NULL);    cout.tie(NULL);

    lint x; cin>>x;

    map<lint, lint> five;
    for (int i=0; i<1000; i++){
        //if (i*i*i*i*i < )
        five[i*i*i*i*i] = i;
    }

    for (int i=0; i<1000; ++i){
        lint need = i*i*i*i*i - x;
        int flag = 1;
        if (need < 0){
            flag = -1;
            need = -need;
        }
        if (five.count(need)){
            cout<<i << " "<<flag*five[need];
            return 0;
        }
    }


}