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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define REP(i,n) for(int i=0; i<(n); i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define ALL(n) begin(n),end(n)
#define IN(a, x, b) (a<=x && x<b)
#define INIT std::ios::sync_with_stdio(false);std::cin.tie(0);
template<class T> inline T CHMAX(T & a, const T b) { return a = (a < b) ? b : a; }
template<class T> inline T CHMIN(T & a, const T b) { return a = (a > b) ? b : a; }

const long double EPS = 1e-10;
const long long INF = 1e18;
const long double PI = acos(-1.0L);

int main()
{
    INIT;
    ll x;
    scanf("%d", &x);
    int a,b;
    for(a = -10e3; a < 10e3; a++){
         for(b = -10e3; b < 10e3; b++){
            if((a*a*a*a*a - b*b*b*b*b) == x)break;
        }
        if((a*a*a*a*a - b*b*b*b*b) == x) break;
    }
    printf("%d %d", a, b);
    return 0;
}