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)

#include<bits/stdc++.h>
#include<iostream>
#include<string>
#include<cmath>
#include<cstdio>
#include<cctype>
#include<cstring>
#include<iomanip>
#include<cstdlib>
#include<ctime>
#include<set>
#include<map>
#include<utility>
#include<queue>
#include<vector>
#include<stack>
#include<sstream>
#include<algorithm>
#define forn(i,a,b)for(int i=(a),_b=(b);i<=_b;i++)
#define fore(i,b,a)for(int i=(b),_a=(a);i>=_a;i--)
#define rep(i,n)for(int i=0,_n=n;i<n;i++)
#define ll long long
#define pii pair<int,int>
#define vi vector<int>
#define vpii vector<pii>
#define m_p make_pair
#define re return
#define pb push_back
#define si set<int>
#define ld long double
#define X first
#define Y second
#define st string
#define ull unsigned long long
#define mod 1000000007
#define INF 1000000007
#define x1 XZVJDFADSPFOE
#define y1 GASDIJSLDAEJF
#define x2 DFDAJKVOHKWIW
#define y2 PSFSAODSXVNMQ
#define LLINF 0x3f3f3f3f3f3f3f3fLL
using namespace std;
inline void read(int &x)
{
	short negative=1;
    x=0;
    char c=getchar();
    while(c<'0' || c>'9')
    {
		if(c=='-')
			negative=-1;
		c=getchar();
	}
    while(c>='0' && c<='9')
        x=(x<<3)+(x<<1)+(c^48),c=getchar();
    x*=negative;
}
ll quickpower(ll n,ll k){
	ll ans=1;
	while(k){
		if(k%2){
			ans*=n;
			ans%=mod;
		}
		n*=n;
		n%=mod;
		k/=2;
	}
	return ans;
}
ll t[311];
int main()
{
	ios::sync_with_stdio(0);
	//think twice,code once
	rep(i,301)
	{
		t[i]=quickpower(i,5);
	}
	ll x;
	cin>>x;
	rep(i,301)
	{
		forn(j,i,300)
		{
			if(t[j]-t[i]==x)
			{
				cout<<j<<" "<<i<<endl;
				return 0;
			}
			if(t[j]+t[i]==x)
			{
				cout<<j<<" "<<-i<<endl;
				return 0;
			}
		}
	}
	return 0;
}