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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define eps 1e-3
#define pi acos(-1.0)
#define inf 0x3f
#define INF 0x3f3f3f3f
#define pb push_back
#define debug1 cout<<"&&";
#define debug2 cout<<"**";
#define ms(a, x) memset(a, x, sizeof(a))
#define for0(i, n) for(int i = 0; i < n; ++i)
#define for1(i, n) for(int i = 1; i <= n; ++i)
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<long long, int> pli;
typedef pair<long long, long long> pll;
const int mod = 1e9 + 7;
const int N = 1e5+10;
const int M = 1e6+10;
/*=================================================================================*/

int a, b, x;
ll Pow(ll k) {
   return k*k*k*k*k;
}

int main() {
   cin>>x;
   for1(i, 1000) {
      ll tmp = Pow(i) -x;
      ll l = -1000, r = 1000;
      while(l<=r) {
         ll mid = (l+r)>>1;
         if(Pow(mid) == tmp) {
            printf("%d %lld\n", i, mid); return 0;
         }
         if(Pow(mid) < tmp) l=mid+1;
         else r=mid-1;
      }
   }
   return 0;
}