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 se second
#define fi first
#define ll long long
#define ld long double
#define pll pair<ll ,ll >
#define plll pair<ll,pll>
#define ppll pair<pll, pll>
#define pii pair<int, int> 
#define pb push_back
#define mod 1000000007
#define N 2000007
ll power(ll a,ll b){
    ll res=1ll;
    while(b>0){
        if(b%2!=0) res=(res*a)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return res%mod;
}
ll ncr(ll n,ll k){
    if(k>n-k) k=n-k;
    ll pro=1;
    for(ll i=0;i<k;i++){
        pro=(pro*(n-i))%mod;
        pro/=(i+1);
    }
    return (pro)%mod;
}

void swap(ll *a,ll *b) {
    ll temp=*a;
    *a=*b;
    *b=temp;
}
vector<ll> prime;
void seive(){
    vector<bool > isprime(1000001,true);
    for(int i=2;i*i<=1000000;i++){
        if(isprime[i]){
            for(int j=i*i;j<=1000000;j+=i){
                isprime[j]=false;
            }
        }
    }
    for(int i=2;i<=1000000;i++){
        if(isprime[i]) prime.push_back(i);
    }
}
vector<ll> parent;
ll find(ll p) {
    if(parent[p]==p) return parent[p];
    return parent[p]=find(parent[p]); 
}
void unun(ll a,ll b) {
    ll r1=find(a);
    ll r2=find(b);
    if(r1!=r2) parent[r2]=r1;
}
void solve() {
    ll n;
    cin>>n;
    if(n==0) {
        cout<<0<<" "<<0<<"\n";
        return ;
    }
    set<ll> dp;
    map<ll,ll> mp;
    for(int i=0;i<4e3;i++) {
        ll val=i*i*i*i*i;
        dp.insert(val);
        mp[val]=i;
    }
    for(int i=0;i>(-4e3);i--) {
        ll val=i*i*i*i*i;
        dp.insert(val);
        mp[val]=i;
    }
    if(dp.find(n)!=dp.end()) {
        ll x=mp[n];
        cout<<x<<" "<<0;
        return ;
    }
    for(auto itr: dp) {
        ll rem=itr-n;
        if(dp.find(rem)!=dp.end()) {
            ll x=mp[rem];
            ll y=mp[itr];
            cout<<y<<" "<<x;
            return ;
        }
    }
    for(int i=0;i<2e6;i++) {
        ll val=i*i*i*i*i;
        ll rem=val-n;
        if(dp.find(rem)!=dp.end()) {
            ll y=mp[rem];
            cout<<i<<" "<<y;
            return;        }
    }
}   


int main(void){
    ios_base::sync_with_stdio(false); 
    cin.tie(NULL);
    cout.tie(NULL);
    int t;
    t=1;
    // cin>>t;
    while(t--) {
        solve();
        // cout<<endl;
    }
    return 0; 

}