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)

#include<algorithm>
#include<iostream>
#include<stdio.h>
#include<iostream>
#include<stdio.h>
#include<vector>
#include<set>
#include<string>
#include<iomanip>
#include<queue>
#include<functional>
#include<cmath>
#include<map>
#include<tuple>
#define rep(i,n) for(int i=0; i<(int)n; i++)//n回繰り返す
#define rrep(i,n)for(int i=n-1;i>=0;i--)
#define REP(i,n) for (int i = 1; i <= (int)n; i++)//i=1からi=nまで
#define RREP(i,n)for(int i=n;i>=1;i--) 
using namespace std;
using ll = long long;
using P = pair<int, int>;
using graph = vector<vector<int>>;
const ll mod = pow(10, 9) + 7;
const int INF = 1e9;
const ll LINF = 1e18;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main() {
    int X; cin >> X;
    rep(a, 110) {
        for (int b = -110; b < 110; b++) {
            if (pow(a, 5) + pow(b, 5) == X) {
                cout << a << " " << b << endl;
                return 0;
            }
        }
    }
}