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<stdlib.h>
#include<math.h>
#include<string.h>
#include<stdbool.h>
#include<assert.h>
typedef long long ll;
typedef long double ld;
#define rep(i,l,r)for(ll i=(l);i<(r);i++)
#define repp(i,l,r,k)for(ll i=(l);i<(r);i+=(k))
#define rrep(i,l,r)for(ll i=(l);i>=(r);i--)
#define INF (1LL<<60)
#define MOD1 1000000007
#define MOD2 998244353
#define MAX_N (1 << 17)
#define YES printf("Yes\n")
#define NO printf("No\n")
#define PN printf("\n")
#define charsize 100005 //10^5+5

void swap(ll *a, ll *b){ll c;c=*b;*b=*a;*a= c;}
// void fill(ll n, ll a[n], ll value){ rep(i,0,n) a[i] = value; }
void cin(ll *n){ scanf("%lld",&(*n)); }
void cin2(ll *a, ll *b){ scanf("%lld%lld",&(*a),&(*b)); }
void cin3(ll *a, ll *b, ll *c){ scanf("%lld%lld%lld",&(*a),&(*b),&(*c)); }
ll max2(ll a,ll b){return a>=b?a:b;}
ll min2(ll a,ll b){return a>=b?b:a;}
ll min3(ll a, ll b, ll c){return (a<=b && a<=c) ? a : b<=c ? b : c;}
ll max3(ll a, ll b, ll c){return (a>=b && a>=c) ? a : b>=c ? b : c;}
ll minn(ll n, ll a[n]){ll b=INF;rep(i,0,n) b=min2(b,a[i]);return b;}
ll maxn(ll n, ll a[n]){ll b=-INF;rep(i,0,n) b=max2(b,a[i]);return b;}
ll ABS(ll a){return a>=0?a:(-a);}
ll POW(ll a, ll b){ll c=1;rep(i,0,b) c*=a;return c;}
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll mod_MOD1(ll n){n+= n<0?((-n)/MOD1+1)*MOD1:0; return n%=MOD1;}
ll mod_p(ll n ,ll p){n+= n<0?((-n)/p+1)*p:0; return n%=p;}
ll change_into_num(char s[] , ll len, ll p){ return !p ? 0 : POW(10,p-1)*(s[len-p]-'0') + change_into_num(s,len,p-1); }

int upll(const void*a, const void*b){return*(ll*)a<*(ll*)b?-1:*(ll*)a>*(ll*)b?1:0;}
int downll(const void*a, const void*b){return*(ll*)a<*(ll*)b?1:*(ll*)a>*(ll*)b?-1:0;}
void sortup(ll*a,int n){qsort(a,n,sizeof(ll),upll);}
void sortdown(ll*a,int n){qsort(a,n,sizeof(ll),downll);}
// int int_sort( const void * a , const void * b ) {
//   return strcmp(( char * )a , ( char * )b );
// }
//     qsort((void*)s,n,sizeof(s[0]),int_sort );

typedef struct{ ll a , b;}fr;
int cmp1( const void *p, const void *q ) { return ((fr*)p) ->a - ((fr*)q)->a;}
int cmp2( const void *p, const void *q ) { return ((fr*)q) ->a - ((fr*)p)->a;}
void strsortup(fr*a,int n){qsort(a,n,sizeof(fr),cmp1);}
void strsortdown(fr*a,int n){qsort(a,n,sizeof(fr),cmp2);}

int main(void){
    // char ;
    ll x;
    // ll a,b;
    cin(&x);
    // cin(&);
    // cin(&);
    // cin2(&,&);
    // cin3(&,&,&);
    // // scanf("%s",);
    // ll [n];
    // rep(i,1,100){
    //     for(ll j=-50;j<i+1;j++){
    //         if( x%(i-j)==0 ){
    //             ll tmp = i*i*i*i ;
    //             printf("%lld %lld ",i,tmp);
    //             tmp +=  i;
    //             if( tmp   ==  x/(i-j) ){
    //                 printf("%lld %lld\n",i,j);
    //                 return 0;
                    
    //                 // - POW(b,2)*POW(a,2) + POW(b,3)*a - POW(b,4)
    //             } 
    //         }
    //     }
    // }
    rep(i,1,1000){
        rep(j,-500,1000){
            if(x== POW(i,5) - POW(j,5) ){
                printf("%lld %lld\n",i,j);
                return 0;
            }
        }
    }

    
    // printf("%lld\n");
    return 0;
}