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> // cout, endl, cin
#include <string> // string, to_string, stoi
#include <vector> // vector
#include <algorithm> // min, max, swap, sort, reverse, lower_bound, upper_bound
#include <utility> // pair, make_pair
#include <tuple> // tuple, make_tuple
#include <cstdint> // int64_t, int*_t
#include <cstdio> // printf
#include <map> // map
#include <queue> // queue, priority_queue
#include <set> // set
#include <stack> // stack
#include <deque> // deque
#include <unordered_map> // unordered_map
#include <unordered_set> // unordered_set
#include <bitset> // bitset
#include <cctype> // isupper, islower, isdigit, toupper, tolower
#include <functional>
#include <numeric>
#define rep(i,n) for(int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main(void) {
	int X;
	cin >> X;
	vector<int> vec(101);
	for (int i = 0; i < 101; i++) {
		vec.at(i) = i * i*i*i*i;
	}
	int A, B;
	for (int i = 0; i < 101; i++) {
		for (int j = 0; j < i; j++) {
			if (X == vec.at(i) - vec.at(j)) {
				A = i;
				B = j;
			}
			if (X == vec.at(i) + vec.at(j)) {
				A = i;
				B = j * -1;
			}
		}
	}
	cout << A << " " << B;
	return 0;
}