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: Runtime Error)

#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 

using namespace std;
using namespace __gnu_pbds; 
  
#define ll long long
#define X first
#define Y second
#define all(x) x.begin(), x.end()
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 

#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 {
#ifdef LOCAL
~debug() { cerr << endl; }
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
};
#define imie(...) " [" << #__VA_ARGS__ ": " << (__VA_ARGS__) << "] "

const int mod = (int)1e9 + 7;
const int MX = (int)1e5 + 10;

void add_self(int &x, int y){x += y; if(x >= mod) x -= mod;}
void sub_self(int &x, int y){x -= y; if(x < 0) x += mod;}
void mul(int &x, int y){x = x * 1LL * y % mod;}

bool ok(ll x){
	ll l = 0, r = 1e3;
	while(l <= r){
		ll md = (l + r) / 2;
		if(md * md * md * md * md <= x) l = md + 1;
		else r = md - 1;
	}
	return r;
}

int main(){

	cin.tie(0); cout.tie(0);
	ios_base::sync_with_stdio(0);

	ll x;
	cin >> x;
	for(ll i = 1; i * i <= x; ++i){
		if(x % i) continue;
		// int j = x / i;
		for(ll a = 0; a <= min(100LL, i); ++a){
			ll b = a - i;
			ll cc = a * a * a * a * a - b * b * b * b * b;
			if(cc == x){
				cout << a << " " << b << endl;
				return 0;
			}
			if(cc == -x){
				cout << -a << " " << -b << endl;
				return 0;
			}
		}
	}

	assert(0);

	return 0;
}