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)

/*
	Powered by C++11.
	Author : Alex_Wei.
*/

#include <bits/stdc++.h>
using namespace std;

//#pragma GCC optimize(3)
//#define int long long

#define ll long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb emplace_back
#define all(x) x.begin(),x.end()
#define sor(x) sort(all(x))
#define rev(x) reverse(all(x))
#define mem(x,v) memset(x,v,sizeof(x))

// Templates start.

namespace IO{
	char buf[1<<21],*p1=buf,*p2=buf,obuf[1<<23],*O=obuf;
	#ifdef __WIN64
		#define gc getchar()
	#else
		#define gc (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?EOF:*p1++)
	#endif
	#define pc(x) (*O++=x)
	#define flush() fwrite(obuf,O-obuf,1,stdout)

	inline int read(){
		int x=0,sign=0; char s=getchar();
		while(!isdigit(s))sign|=s=='-',s=getchar();
		while(isdigit(s))x=(x<<1)+(x<<3)+(s-'0'),s=getchar();
		return sign?-x:x;
	}
	void print(int x) {if(x>9)print(x/10); pc(x%10+'0');}
}

namespace math{
	const int mod=998244353;
	const int maxn=1e6+5;

	ll ksm(ll a,ll b){ll s=1,m=a; while(b){if(b&1)s=s*m%mod; m=m*m%mod,b>>=1;} return s;}
	ll inv(ll x){return ksm(x,mod-2);}

	ll fc[maxn],ifc[maxn];
	void init_m(int n){
		fc[0]=1; for(int i=1;i<=n;i++)fc[i]=fc[i-1]*i%mod;
		ifc[n]=inv(fc[n]); for(int i=n-1;i>=0;i--)ifc[i]=ifc[i+1]*(i+1)%mod;
	}
	ll C(ll n,ll m){return fc[n]*ifc[m]%mod*ifc[n-m]%mod;}
}

// Templates end. Please give me more points.

using namespace IO;
//using namespace math;


ll x;

int main(){
	cin>>x;
	for(ll i=1;i<=3e3+5;i++){
		ll tmp=i*i*i*i*i-x;
		ll l=-i,r=i;
		while(l<r){
			ll m=(l+r>>1)+1;
			if(m*m*m*m*m<=tmp)l=m;
			else r=m-1;
		}
		if(l*l*l*l*l==tmp)cout<<i<<" "<<l<<endl,exit(0);
	}
	return 0;
}