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 <stdio.h>
#include <vector>
#include <map>
#include <unordered_map>
#include <set>
#include <unordered_set>
#include <stack>
#include <queue>
#include <algorithm>
#include <string.h>
#include <string>
#include <math.h>
#include <iomanip>
using namespace std;

#define SORT(v) sort((v).begin(), (v).end())
#define RSORT(v) sort((v).rbegin(), (v).rend())
#define pb push_back

typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;

const int mod = (int)1e9 + 7;
const int mod2 = 998244353;

ll exp(ll taban, ll us) {
    ll carpan = taban % mod;
    ll temp = us;
    ll res = 1;
    while(temp){
        if(temp % 2) res = (res*carpan) % mod;
        temp /= 2;
        carpan = (carpan*carpan) % mod;
    }
    return res;
}
 
ll ebob(ll a, ll b){
    if(!a)return b;
    return ebob(b%a, a);
}

ll ekok(ll a, ll b){
    return (a*b)/ebob(a, b);
}
ll fonk(ll n){
	return n*n*n*n*n;
}
int main(){
    ios_base::sync_with_stdio(false); cin.tie(NULL);
	ll x; cin>>x;
	for(ll i = -1e3; i <= 1e3; i++){
		for(ll j = -1e3; j <= 1e3; j++){
			if(fonk(j) - fonk(i) == x){
				cout<<j<<" "<<i<<endl;
				return 0;
			}
		}
	}
}