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>
using namespace std;
#define rep(i, n) for(int i=0; i<n; i++)
#define rep1(i, n) for(int i=1; i<=n; i++)
#define repr(i, n) for(int i=n-1; i>=0; i--)
#define repr1(i, n) for(int i=n; i>=1; i--)
#define all(v) v.begin(),v.end()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int INF = 1e9;
const ll LLINF = 1e18;
const ll MOD = 1e9+7;
const double EPS = 1e-10;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

ll fifth(ll x) {
    return x * x * x * x * x;
}

int main() {
    ll x; cin >> x;
    for (ll a = -100; a <= 100; a++) {
        for (ll b = -100; b <= 100; b++) {
            if (fifth(a) - fifth(b) == x) {
                cout << a << " " << b << endl;
                return 0;
            }
        }
    }
}