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 <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>

#include <vector>
#include <cstring>
#include <queue>
#include <map>
#include <set>

#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>

using namespace std;

typedef vector<int> vi; 
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef long long ll;
#define sz(a) int((a).size()) 
#define pb push_back 
#define all(c) (c).begin(),(c).end() 
#define rep(i, n) for(int i = 0; i < n; ++i) 
#define foreach(it, c) for(__typeof((c).begin()) it = (c).begin(); it != (c).end(); ++it) 

#define TRACE(x...) x
#define watch(x) TRACE(cout << #x" = " << x << endl)
#define watcharr(i, x) TRACE(cout << #x" = "); rep(i, sz(x)) cout << x[i] << " "; cout << endl

int main() {
	
	ifstream infile("input.in");
	if(infile.good()) cin.rdbuf(infile.rdbuf());
	
    int X; cin >> X;

    vector<ll> v;
    ll a = 0, k = 0;
    while(k < X) {
        k = pow(a++, 5.0);
        v.pb(k);
    }

    int ra = 0, rb = 0;
    rep(i, sz(v)) {
        ll b5 = v[i] - X;
        rep(j, sz(v)) 
            if(v[j] == abs(b5))
                ra = i, rb = j * (b5 >= 0 ? 1 : -1);
    }

    cout << ra << " "  << rb << endl;

	if(infile.good()) infile.close();
	return 0;
	
}