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;
// -----------------<fast IO>------------------
#define IOS             ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define FLSH             fflush(stdout)
#define fileIO(name) \
    freopen(name".in", "r", stdin); \
    freopen(name".out", "w", stdout);
#define PRECISION(x)     cout << fixed << setprecision(x);
#define endl             "\n"
// -----------------<Container manipulation / traversal macros>------------------
#define mp                 make_pair
#define pb                 push_back
#define ff                 first
#define ss                 second
#define sz(x)             ((int)((x).size()))
#define all(x)             (x).begin(),(x).end()
#define rep(i,n)         for(ll (i)=0;(i)<(n);(i)++)
#define forn(i,a,b)     for(ll (i)=(a);(i)<=(b);(i)++)
#define fill(l,a)        memset((l) , (a) , sizeof(l))
// -----------------<Some common useful functions>------------------
#define maX(a,b)        ( (a) > (b) ? (a) : (b))
#define miN(a,b)        ( (a) < (b) ? (a) : (b))
#define checkbit(n,b)   ( (n >> b) & 1)
#define INDEX(arr,i)      ( lower_bound(all(arr),i)-arr.begin())
#define in(b,a)         ( (b).find(a) != (b).end())
// -----------------<TypeDef Start>------------------
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll> pii;
typedef vector<ll> vi;
typedef vector< std::vector<ll> > matrix;
typedef map<ll,ll> mii;
typedef vector<pii> vpii;
// -----------------<Constants>------------------
const int MAXN = 1000005;
const int SQRTN = 1003;
const int LOGN = 22;
const double PI=acos(-1);

const int INF=1e9;

int MOD = 1000000007;
const int FMOD = 998244353;
const double eps = 1e-9;
ll gcd(ll a,ll b)
{
    if(a==0)
    return(b);
    return(gcd(b%a,a));
}
ll lcm(ll a,ll b)
{
    return(a/gcd(a,b)*b);
}
/*ll power(ll a,ll b,ll p)
{
    ll res=1;
    a%=p;
    while(b>0)
    {
        if(b&1)
        res=res*a%p;
        a=a*a%p;
        b>>=1;
    }
    return(res);
}
ll modInverse(ll a, ll m)
{
    ll g = gcd(a, m);
    return(power(a,m-2,m));
}*/
/*ll b_search(ll arr[],ll n,ll s)
{
    ll l=0,r=n-1;
    while(l<=r)
    {
        ll mid=l+(r-l)/2;
        if(arr[mid]==s)
        return(mid);
        if(arr[mid]<s)
        l=mid+1;
        else
        r=mid-1;
    }
    return(-1);
}*/
bool isprime(ll x)
{
    if(x==1)
        return(false);
    for(ll i=2;i*i<=x;i++)
    {
        if(x%i==0)
            return(false);
    }
    return(true);
}
//vector<ll> v[100005];
void solve()
{
    ll n;
    cin>>n;
    for(ll i=-155;i<=155;i++)
    {
        for(ll j=-155;j<=155;j++)
        {
            if((pow(i,5)-pow(j,5))==n)
            {
                cout<<i<<" "<<j<<endl;
                return;
            }
        }
    }
}

int main()
{
    IOS
    ll t=1;
    //cin>>t;
    while(t--)
    {
        solve();
    }
    return(0);
}