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)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#include <limits.h>

#define max  100000

//qsort(str, n, sizeof(int), up_comp);
int up_comp(const void *a, const void *b){return *(int*)a - *(int*)b;}
int down_comp(const void *a, const void *b){return *(int*)b - *(int*)a;}
int count_digit(int a){int i=0;while(a!=0){a=a/10;i++;}return i;}
int euclid( int a, int b ){int temp;if(a<b){temp=a;a=b;b=temp;}if(b<1)return -1;if(a%b==0) return b;return euclid(b,a%b);}
int int_sort( const void * a , const void * b ) {
  return strcmp(( char * )a , ( char * )b );
}

int main()
{
    int x, a, b;
    int i, j, cnt=0;
    
    scanf("%d", &x);
    
    for (b=-1000; b<=1000; b++) {
        for (a=-1000; a<=1000; a++) {
            if (pow(a, 5)-pow(b, 5)==x) {
                printf("%d %d", a, b);
                return 0;
            }
        }
    }
    
    return 0;
}