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)

// SeeAlso: 

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

#define MAX 100000
#define NIL -1
#define MOD 1000000007

typedef int _loop_int;
#define rep(i,n) for(int i = 0; i < n; i++)
#define FOR(i,a,b) for(_loop_int i=(_loop_int)(a);i<(_loop_int)(b);++i)
#define FORR(i,a,b) for(_loop_int i=(_loop_int)(b)-1;i>=(_loop_int)(a);--i)

#define debug(x) cout<<#x<<": "<<x<<endl
#define debig_vec(v) cout<<#v<<":";rep(i,v.size())cout<<" "<<v[i];cout<<endl
#define ALL(a) (a).begin(),(a).end()

// 最大公約数
inline constexpr ll gcd(ll a,ll b){if(!a||!b)return 0;while(b){ll c=b;b=a%b;a=c;}return a;}
// 最小公倍数
inline constexpr ll lcm(ll a,ll b){if(!a||!b)return 0;return a*b/gcd(a,b);}

#define print2D(h, w, arr) rep(i, h) { rep(j, w) cout << arr[i][j] << " "; cout << endl; }
template<class T> void print(const T& x){cout << setprecision(12) << x << endl;}
template<class T, class... A> void print(const T& first, const A&... rest) { cout << first << " "; print(rest...); }

int main() {
    ll x;
    cin >> x;
    
    ll maxm = 2000;

    for(ll i = (-maxm);i<=maxm;i++) {
        ll a = 1;
        rep(aaaa, 5) {
            a *= i;
        }
        ll left = -maxm;
        ll right = maxm;

        ll resultB;
        while(left<right) {
            ll middle = (right+left)/2;
            ll b = 1;
            rep(aaaa, 5) {
                b *= middle;
            }

            if ((a-b)==x) {
                cout << i << " " << middle << endl;
                return 0;
            }            

            if((a-x) < b) {
                right = middle-1;
            } else {
                left = middle+1;
            }

            resultB = middle;
        }

        ll b = 1;
        rep(aaaa, 5) {
            b *= resultB;
        }
    }
    
    return 0;
}