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"

using namespace std;

#define rep(i,n) for(int i=0;i<(int)(n);i++)
void sort2array(double *a,long *b,long n){
    for(int i=0;i<n;i++){
        b[i]=i;
    }
    sort(b,b+n,[a](long i,long j)->bool {return a[i]<a[j];} );
}
//bを何回足せばaを超えるか
int wtover(int a,int b){
    if(a%b>0)return (a/b)+1;
    else return a/b;
}
const int INF=100000001,ID_MAX=20;
int bi_e[ID_MAX]={0};
//2進数表示したときの最高桁
int bi_max(long n){
    int m=0;
    for(m;pow(2,m)<=n;m++);m=m-1;
    return m;
}
//bi_eに二進数表示したやつを代入
void bi_exs(long n){
    rep(i,ID_MAX){
        bi_e[i]=0;
    }
    if(n<pow(2,ID_MAX)){
        int m=bi_max(n);
        long l;
        for(m;m>-1;m--){
            l=pow(2,m);
            bi_e[m]=n/l;
            n=n%l;
        }
    }
}
//x^n mod m (nが負の時は０)
long myPow(long x, long n, long m){
  if(n<0)return 0;
  if(n == 0)
    return 1;
  if(n % 2 == 0)
    return myPow(x * x % m, n / 2, m);
  else
    return x * myPow(x, n - 1, m) % m;
}


int main(){
    long int i,n,x,y;
    //入力
    cin>>n;
    //処理
    bool tof=false;
    /*for(int i=1;i<=floor(sqrt(n));i++){
        if(n%i==0){
            for(int j=-109;i<=109;i++){
                long int a=j+i,b=j;
                if(pow(a,5)-pow(b,5)==n){
                    x=a;
                    y=b;
                    tof=true;
                    break;
                }
            }
        }
        if(tof)break;
    }*/
    for(int i=-109;i<=109;i++){
        for(int j=-109;j<=109;j++){
            long re=pow(i,5)-pow(j,5);
            if(re==n){
                x=i;
                y=j;
                tof=true;
                break;
            }
        }
        if(tof)break;
    }
    //出力
    cout<<x<<" "<<y<<endl;
}