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: Time Limit Exceeded)

#include <iostream>
#include <iomanip>
#include <string>
#include <stack>
#include <vector>
#include <math.h>
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <iterator>
#include <map>
#include <set>
#include <queue>
#include <list>
#include <regex>
#include <time.h>
#include <complex>
#include <cstdint>
using namespace std;
using pii = pair<int, int>;
using ll = long long;
using ld = long double;
#define pb push_back
#define mp make_pair
#define sc second
#define fr first
#define stpr setprecision
#define cYES cout<<"YES"<<endl
#define cNO cout<<"NO"<<endl
#define cYes cout<<"Yes"<<endl
#define cNo cout<<"No"<<endl
#define rep(i,n) for(ll i=0;i<(n);++i)
#define Rep(i,a,b) for(ll i=(a);i<(b);++i)
#define rrep(i,n) for(int i=n-1;i>=0;i--)
#define rRep(i,a,b) for(int i=a;i>=b;i--)
#define crep(i) for(char i='a';i<='z';++i)
#define psortsecond(A,N) sort(A,A+N,[](const pii &a, const pii &b){return a.second<b.second;});
#define ALL(x) (x).begin(),(x).end()
#define debug(v) cout<<#v<<":";for(auto x:v){cout<<x<<' ';}cout<<endl;
#define endl '\n'
#define mod 1000000007
typedef long long ll;

/*
unsigned gcd(int a, int b) {
	while (1) {
		if (a < b) swap(a, b);
		if (!b) break;
		a %= b;
	}
	return a;
}


ll comb(int n, int r) {
	if (r * 2 > n) r = n - r;
	uintmax_t dividend = 1;
	uintmax_t divisor = 1;
	for (unsigned int i = 1; i <= r; ++i) {
		dividend *= (n - i + 1);
		divisor *= i;
	}
	return dividend / divisor;
}

int compare_char(const void* left, const void* right) {
	char *left_char = (char *)left;
	char *right_char = (char *)right;

	return strcmp(left_char, right_char);
}
*/


int main(){


//	ios::sync_with_stdio(false);
//	cin.tie(0);

	//int N, M, *A, *B, i, j, *W, *flag, sum=0, start, end, length, tmp;
	ll i, j, k, l, m, n, len;
	ll C, D, N, M, K, num_R, num_G, num_B, temp;
	ll sum = 0, sum_B = 0, sum_C = 0, max, min, count = 0, Q, W, X, flag=0;
	ll  A, B, *H, *ans;



	ll temp1, temp2;

	cin >> X;

	i = pow(X, 0.2);

	while (1){
		flag = 0;
		A = i;
		B = A*A*A*A*A - X;
		if (B < 0)
			flag = 1;

		B = abs(B);

		j = pow(B, 0.2);

		if (B == j*j*j*j*j){
			if (flag == 1)
				j *= -1;

			cout << A << " " << j << endl;
			break;
		}
	}


	return 0;

}//1 2 3 4 5 6