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: Accepted)

#include <bits/stdc++.h>
using namespace std;


typedef long long ll;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector<ii> vii;
typedef long double ld;

typedef set<int>::iterator sit;
typedef map<int,int>::iterator mit;
typedef vector<int>::iterator vit;

const int INF = 1e9 + 7;
const int MOD = 1e9 + 7;
const int MAXN = 1e6 + 3;

#define _  %  MOD
#define __ %= MOD

#define      each(it,s)        for(auto it = s.begin(); it != s.end(); ++it)
#define      sortA(v)          sort(v.begin(), v.end())
#define      sortD(v)          sort(v.begin(), v.end(), greater<auto>())
#define      fill(a)           memset(a, 0, sizeof (a))

#define      swap(a,b)        {a=a+b;b=a-b;a=a-b;}
#define      rep(i, n)         for(ll i = 0; i < (n); ++i)
#define      repA(i, a, n)     for(ll i = a; i <= (n); ++i)
#define      repD(i, a, n)     for(ll i = a; i >= (n); --i)
#define watch(x) cout << (#x) << " is " << (x) <<endl

#define fi first
#define se second
#define mp make_pair
#define pb push_back

#define fbo find_by_order
#define ook order_of_key

ll gcd(ll a,ll b){if(b==0) return a; else return gcd(b,a%b);}
ll power(ll x,ll y){ll res=1; while(y>0){if(y&1) res=res*x;y=y>>1;x=x*x;}return res;} // modular exponent

int main()
{
    ios_base::sync_with_stdio(false); // don't use printf and scanf
    cin.tie(NULL); // cout<<fixed<<setprecision
    ll x;
    cin>>x;
    int a,b;
    for(a=-1000;a<=1000;a++)
    {
        for(b=-1000;b<=1000;b++)
        {
            ll t1=power(a,5ll);
            ll t2=power(b,5ll);
            if((t1-t2)==x)
            {
                cout<<a<<" "<<b<<'\n';
                return 0;
            }
        }
    }

}
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