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: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <iomanip>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstring>
#include <climits>

using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define REP(i, n) FOR(i, 0, n)
#define SORT(c) sort((c).begin(), (c).end())

typedef long long ll;
typedef pair<int, int> P;
typedef vector<int> V;
typedef map<int, int> M;

constexpr ll INF = 1e18;
constexpr ll MOD = 1e9 + 7;
constexpr double PI = 3.14159265358979323846;
constexpr int di[] = {0, 0, 1, -1};
constexpr int dj[] = {1, -1, 0, 0};

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);

    ll x;
    cin >> x;

    int res = 0;

    for (ll a = -500; a <= 500; a++)
    {
        for (ll b = -500; b <= 500; b++)
        {
            if (a * a * a * a * a - b * b * b * b * b == x)
            {
                cout << a << " " << b << endl;
                return 0;
            }
        }
    }

    return 0;
}