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)

/*
██╗  ██╗ █████╗ ██████╗ ███████╗███████╗███╗   ███╗        ██╗  ██╗██╗  ██╗ █████╗ ██╗     ███████╗██████╗
██║ ██╔╝██╔══██╗██╔══██╗██╔════╝██╔════╝████╗ ████║        ██║ ██╔╝██║  ██║██╔══██╗██║     ██╔════╝██╔══██╗
█████╔╝ ███████║██████╔╝█████╗  █████╗  ██╔████╔██║        █████╔╝ ███████║███████║██║     █████╗  ██║  ██║
██╔═██╗ ██╔══██║██╔══██╗██╔══╝  ██╔══╝  ██║╚██╔╝██║        ██╔═██╗ ██╔══██║██╔══██║██║     ██╔══╝  ██║  ██║
██║  ██╗██║  ██║██║  ██║███████╗███████╗██║ ╚═╝ ██║███████╗██║  ██╗██║  ██║██║  ██║███████╗███████╗██████╔╝
╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═╝     ╚═╝╚══════╝╚═╝  ╚═╝╚═╝  ╚═╝╚═╝  ╚═╝╚══════╝╚══════╝╚═════╝
				Believe in yourself ,( try and try and then try ), You can do it  
*/
#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
#include<unordered_map>
#define endl "\n"
#define Ceil(x,y) ((x+y-1)/y)
#define clr(arr, x) memset(arr, x, sizeof arr)
#define all(v) v.begin(),v.end()
#define allr(s) s.rbegin(),s.rend()
#define rt(s) return cout<<s,0
#define watch(x) cout<<(#x)<<" = "<<x<<endl
#define sz(s)	(int)(s.size())
#define OO 0x3f3f3f3f3f3f3f3fLL
const int oo = 0x3f3f3f3f, mod = 1e9 + 7;
using namespace std;
typedef long long ll;
int dx[] { 1, -1, 0, 0, 1, 1, -1, -1 };
int dy[] { 0, 0, 1, -1, 1, -1, 1, -1 };
void K_K(){ 
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
}

int main(){
	K_K();
	double x;
	cin >> x;
	ll l = -1e3, r = 1e3;
	pair<ll, ll>ans;
	while (l <= r){
		ll a = (l + r) >> 1;
		ll b = pow(a, 5) - x;
		ll eq = pow(a, 5) - pow(b, -5);
		if (eq == x){
			ans = { a, b };
			break;
		}
		else if (eq < x)
			l = a + 1;
		else
			r = a - 1;
	}
	cout << ans.first << ' ' << ans.second;
}