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)

//#pragma GCC optimize ("Ofast")
//#pragma GCC optimize ("unroll-loops")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//#include <ext/rope>
#define pb push_back
#define F first
#define S second
#define ins insert
#define mp make_pair
#define fo(i, n1, n, x) for(int i = n1; i <= n; i += x)
#define foo(i, n, n1, x) for(int i = n; i >= n1; i -= x)
#define bit __builtin_popcount
#define md (l + ((r - l) / 2))
#define all(x) x.begin(),x.end()
#define eb emplace_back
#define ub upper_bound
#define lb lower_bound
#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define file(s) if (fopen(s".in", "r")) freopen(s".in", "r", stdin), freopen(s".out", "w", stdout)

using namespace std;
//using namespace __gnu_pbds;
//using namespace __gnu_cxx;


using ll = long long;

#define int ll

using pii = pair<int,int>;
using pll = pair<ll,ll>;
using vi = vector<int>;



const int N = 2e5 + 11, mod = 1e9 + 7, mod2 = 998244353;
const int MAX = 1e5 + 11;
const int INF1 = 2e9 + 11;
const ll INF2 = 6e18 + 11;
const double INF3 = 1e8 + 11;
const int base = 500;
const int P = 31;
const int dx[] = {1, -1, 0, 0, 1, 1, -1, -1};
const int dy[] = {0, 0, 1, -1, 1, -1, 1, -1};
const double EPS = 1e-4;
const double PI = acos(-1.0);


//template<typename T> using ordered_set = tree <T, null_type, less <T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T1, typename T2> inline void chmin(T1 &a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1 &a, T2 b) { if (a < b) a = b; }


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int x;
int root(int x) {
  int l = 0, r = cbrtl(x);
  while (l <= r) {
    if (md * md * md * md * md <= x) l = md + 1;
    else r = md - 1;
  }
  return l - 1;
}
main() {
  file("threesum");
  ios;
  cin >> x;
  ll a = 5e18;
  a = root(a);
  fo(i, 1, a, 1) {
    int c = i * i * i * i * i - x;
    int d = root(c);
    if (d * d * d * d * d + x == c + x) return !(cout << i << " " << d);
  }
  return 0;
}
/**
a^5 = x + b^5
*/