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)

#pragma GCC optimize("O3")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds; 

#define int long long
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define all(v) (v).begin(),(v).end()
#define rall(v) (v).rbegin(),(v).rend()
#define f first
#define s second
#define sz(x) (int)x.size()
#define endl '\n'
#define forn(i,n) for(int i=0;i<n;++i)
#define fore(i,l,r) for(int i=int(l);i<=int(r);++i)
#define rep(i,begin,end) for(__typeof(end) i=(begin);i!=(end);i++)
#define fill(a,value) memset(a,value,sizeof(a));
#define gcd(a,b) __gcd((a),(b))
#define watch1(x) cout << (x) << endl
#define watch2(x,y) cout << (x) << " " << (y) << endl
#define watch3(x,y,z) cout << (x) << " " << (y) << " " << (z) << endl
#define fastio ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef long double ld; 
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef tree<pii,null_type,less<pii>,rb_tree_tag,tree_order_statistics_node_update> oset;

// find_by_order(k) returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;

const int mod = 1e9 + 7;
const int inf = 9e18;
const int N = 1e5 + 5;

int i,x,a,b;

void solve(){
    cin>>x;
    map<int,int> mp;
    for(i=1;i*i*i*i*i<inf;++i){
        mp[i*i*i*i*i]=i;
    }
    for(i=1;i*i*i*i*i<inf;++i){
        if(mp[x+i*i*i*i*i]>0){
            cout<<mp[x+i*i*i*i*i]<<" "<<i;
            return;
        }
    }
    for(i=0;;--i){
        if(mp[x+i*i*i*i*i]>0){
            cout<<mp[x+i*i*i*i*i]<<" "<<i;
            return;
        }
    }
}

signed main(){
    fastio;
    int t;
    //cin>>t;
    t=1;
    while(t--){
        solve();
        cout<<endl;
    }
    return 0;
}