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)

//#pragma GCC optimize("Ofast")

#pragma GCC optimize("O3")
//#pragma GCC target("avx,avx2,fma")
#include<iostream>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstdlib>
#include<set>
#include<vector>
#include<sstream>
#include<queue>
#include <iomanip>
#include<unordered_set>
#include <stack>
#include <unordered_map>
#include <map>
constexpr auto INF = 9223372036854775807;
typedef long long int ll;
typedef unsigned long long int ull;
typedef unsigned long int ul;
#define f(i,a,b) for(ll i=(ll)a;i<(ll)b;i+=1)
#define rf(i,a,b) for(ll i=(ll)a;i>=(ll)b;i-=1)
#define endl '\n'
#define N 1000000007// prime modulo value 
#define C 1000000001
#define M 998244353
#define all(x) x.begin(),x.end()
using namespace std;
// From Geeksforgeeks
inline ll gcd(ll a, ll b)
{
	if (b == 0)
		return a;
	return gcd(b, a % b);
}
// Power function took from geeksforgeeks
ll power(ll x, ll y)
{
	ll res = 1;

	x = x;
	if (x == 0) return 0;
	while (y > 0)
	{
		if (y & 1)
			res = (res * x);
		y = y >> 1;
		x = (x * x);
	}
	return res;
}
double dist(ll x1, ll y1, ll x2, ll y2)
{
	double ans = sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
	return ans;
}
int main()
{
	ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	ll x;
	cin >> x;
	f(i, 0, C)
	{
		char flag = 'N';
		f(j, -i-1, i + 2)
		{
			if (power(i, 5) - power(j, 5) == x)
			{
				cout << i << " " << j;
				flag = 'Y';
				break;
			}
		}
		if (flag == 'Y')
			break;
	}
	return 0;
}