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 <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack> 
#include <set>
#include <map>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <cassert>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
using edge = struct {ll to; ll cost;};
using point = struct {ll x; ll y;};
typedef string str;
typedef std::pair<ll,ll> pl;
typedef std::pair<ll,pl> pl3;
typedef std::map<string, ll> msl;
typedef std::map<char, ll> mcl;
typedef std::map<ll, ll> mll;
typedef std::vector<ll> vl;
typedef std::vector<pl> vpl;
typedef std::vector<point> points;
typedef std::vector<pl3> vpl3;
typedef std::priority_queue<ll> pq;
typedef std::priority_queue<ll, vl, greater<ll>> pql; // priority queue taking from the lower value.
typedef std::vector<edge> gr;
const ll MOD = 1e9 + 7;
// const ll MOD = 998244353;
const ll INF = MOD * MOD;
const long double EPS = 1e-9;
const long double PI = 3.14159265358979323846;
vpl dirs = {{1, 0}, {0, 1}, {-1, 0}, {0, -1},  // four directions
            {1, 1}, {-1, 1}, {-1, -1}, {1, -1},  // diagonal
            {0, 0}  // self
            };

#define rep(i, n) for(ll (i) = 0; (i) < (n) ; (i)++ )
#define revrep(i, n) for(ll (i) = n - 1; (i) >= 0 ; (i)--)
#define For(i, a, b) for(ll (i) = (a); (i) < (b) ; (i)++ )
#define revFor(i, b, a) for(ll (i) = (b) - 1; (i) >= (a) ; (i)--)
#define all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define isUpper(c) ('a' - c > 0)
#define isNum(c) (0 <= (c) - '0' && (c) - '0' <= 9)
#define toLower(c) char((c)+0x20)
#define toUpper(c) char((c)-0x20)
#define pb push_back
#define mp make_pair
#define pr(a) cout << (a)
#define prl(a) cout << (a) << endl
#define prl2(a,b) cout << (a) << " " << (b) << endl
#define prl3(a,b,c) cout << (a) << " " << (b) << " " << (c) << endl
#define prl4(a,b,c,d) cout << (a) << " " << (b) << " " << (c) << " " << (d) << endl
#define prs(a) cout << (a) << " "
#define yn(condition) if ((condition)) prl("Yes"); else prl("No");
#define YN(condition) if ((condition)) prl("YES"); else prl("NO");
#define in1(a) cin >> (a)
#define in2(a,b) cin >> (a) >> (b)
#define in3(a,b,c) cin >> (a) >> (b) >> (c)
#define in4(a,b,c,d) cin >> (a) >> (b) >> (c) >> (d)
#define e1 first
#define e2 second
#define ctol(c) ll((c)) - ll('0')
#define ltos(n) to_string((n))
#define items(kv, v) for(auto &(kv): (v))
#define ndig(N, n) ctol(ll(ltos((N))[ll(ltos((N)).length()) - (n)]))
#define rsort(a,n) sort(a,a+n,greater<>())
#define Forchar(c,a,z) for(char (c) = (a); (c) <= (z); (c)++)
#define cntchar(s,c) count(all((s)), c)
#define substring(s, start, end) s.substr((start), (end) - (start) + 1)
#define prl_nd(num, digits) cout << fixed << setprecision(digits) << (num) << endl;
#define XOR(a, b) (a)^(b)
#define prl_time(s) prl3("Elapsed Time:", 1000.0 * (clock() - s) / CLOCKS_PER_SEC, "[ms]");
#define char_to_str(c) string(1, (c))
#define DEBUG(x) cerr<<#x<<": "<<(x)<<endl
#define DEBUG_VEC(v) cerr<<#v<<": ";rep(__i,(v).size())cerr<<((v)[__i])<<", ";cerr<<endl

struct MaxFlow {
    struct F_edge {
        ll to, rev, capacity;
        F_edge(ll to, ll rev, ll capacity) : to(to), rev(rev), capacity(capacity) {}
    };

    typedef vector<F_edge> F_edges;
    vector<F_edges> graph;
    ll n_vertex;
    // level is the shortest path to get a given node from the source node.
    vl level, iter;

    MaxFlow(ll n_vertex) : n_vertex(n_vertex) {graph.resize(n_vertex);}

    void add_edge(ll from, ll to, ll capacity) {
        graph[from].pb({to, ll(graph[to].size()), capacity});
        graph[to].pb({from, ll(graph[from].size()) - 1, 0});
    }

    void bfs(ll source){
        level = vl(n_vertex, -1); level[source] = 0;
        queue<ll> q; q.push(source);
        while (!q.empty()) {
            ll vertex = q.front(); q.pop();
            rep(i, graph[vertex].size()){
                ll target = graph[vertex][i].to;
                ll cap_target = graph[vertex][i].capacity;
                // if the flow can be into the target node, implement below.
                if (cap_target > 0 && level[target] < 0){
                    level[target] = level[vertex] + 1;
                    q.push(target);
                }
            }
        }
    }

    ll dfs(ll vertex, ll sink, ll flow) {
        if (vertex == sink) return flow;
        for(ll &i = iter[vertex]; i < graph[vertex].size(); i++){
            ll target = graph[vertex][i].to; ll cap_target = graph[vertex][i].capacity;
            ll rev_target = graph[vertex][i].rev;
            // if capasitiy is not full yet and target is farther,
            // then assign current flow 
            if (cap_target > 0 && level[vertex] < level[target]) {
                ll d = dfs(target, sink, min(cap_target, flow));
                if (d > 0){ // if the flow successfully reaches the sink, reduce the flow from the capacity 
                    graph[vertex][i].capacity -= d;
                    graph[target][rev_target].capacity += d;
                    return d;
                }
            }
        }
        return 0;
    }

    ll dinic(ll source, ll sink) {
        ll flow = 0;
        while (true) {
            bfs(source);
            // if there is no path leading to the sink, the maximum flow is 0.
            if (level[sink] < 0) return flow;
            iter = vl(n_vertex, 0); ll f;
            while ((f = dfs(source, sink, INF)) > 0) flow += f;
        }
    }
};

class UnionFind {
    vl parents, set_size;
public:
    UnionFind(){}
    UnionFind(ll n){
        parents = set_size = vl(n);
        rep(i, n) {parents[i] = i; set_size[i] = 1LL;}
    }

    ll root_find(ll x){
        if (parents[x] == x) return x;
        return parents[x] = root_find(parents[x]);
    }

    void unite(ll x, ll y){
        x = root_find(x); y = root_find(y);
        if (x == y) return;
        if (set_size[x] < set_size[y]){
            parents[y] = x; set_size[x] += set_size[y];
        }else{
            parents[x] = y; set_size[y] += set_size[x];
        }
    }

    bool is_same(ll x, ll y) {// connected or not
        return root_find(x) == root_find(y);
    }

    ll size(ll x) {
        return set_size[root_find(x)];
    }
};

/*
class LCA{
public:
    ll N, logN;
    vl depth, len;
    gr tree[200005]; // global declaration later.
    vector<vl> parents;

    LCA(ll n){
        N = n;
        logN = 0;
        while (N > (1LL << logN)) logN++;
        depth = vl(N); len = vl(N);
        parents = vector<vl>(logN, vl(N));
        init(0, -1, 0, 0);
        build();
    }

    void init(ll source, ll parent, ll d, ll l){
        depth[source] = d;
        parents[0][source] = parent;
        len[source] = l;
        rep(i, tree[source].size()){
            ll target = tree[source][i].to;
            ll cost = tree[source][i].cost;
            if (target == parent) continue;
            init(target, source, d + 1, cost + l);
        }
    }

    void build(){
        rep(k, logN - 1) rep(n, N){
            // if there is no parent, -1.
            // otherwise, the parent of the parent is the parent.
            if (parents[k][n] < 0) parents[k + 1][n] = -1;
            else parents[k + 1][n] = parents[k][parents[k][n]];
        }
    }

    ll query(ll u, ll v){
        if (depth[u] > depth[v]) swap(u, v);
        rep(k, logN) if ((depth[v] - depth[u]) >> k & 1) v = parents[k][v];
        if (u == v) return u;

        revrep(k, logN){
            if (parents[k][u] != parents[k][v]){
                u = parents[k][u]; v = parents[k][v];
            }
        }
        return parents[0][u];
    }

    ll distance(ll u, ll v){
        ll w = query(u, v);
        return len[u] + len[v] - 2 * len[w];
    }

};
*/

struct BIT{
    ll n; vl tree_dat;
    BIT(ll n): tree_dat(n + 1, 0), n(n){};

    //       x: 1001 1010  1100 1011 1101  1111
    // x & - x: 0001 0010  0100 0001 0001  0001
    //      ->: 1010 1100 10000 1100 1100 10000

    ll update_func(ll val, ll dat){
        // if maximum -> max(val, dat)
        // return max(val, dat);
        // if cumulative sum
        return val + dat;
    }

    ll query(ll i){
        /*
        e.g.) i = 10101
        itr1. 10101 -> 10100 
        itr2. 10100 -> 10000
        itr3. 10000 -> 00000 (break)
        */
        ll ret = 0;
        for (ll j = i; j >= 0; j = (j & (j + 1)) - 1){
            ret = update_func(ret, tree_dat[j]);
        }
        return ret;
    }

    ll lower_bound(ll key){
        if (key <= 0) return 0;
        ll left = 0, right = 1;
        while(right <= n) right *= 2;

        for (ll i = right; i > 0; i /= 2){
            if (left + i <= n && tree_dat[left + i - 1] < key){
                key -= tree_dat[left + i - 1];
                left += i;
            }
        }
        return left;
    }

    void update(ll i, ll val){
        /*
        e.g.) i = 10101, n = 11111
        itr1. 10101 ->  10110 
        itr2. 10110 ->  11000
        itr3. 11000 -> 100000 (break)
        */
        if (i < 0) return;
        for(ll j = i; j < n; j |= j + 1){
            tree_dat[j] = update_func(val, tree_dat[j]);
        }
    }
};

ll gcd(ll m, ll n){
    ll a = max(m, n);
    ll b = min(m, n);
    while(b != 1 && b != 0){ a %= b; swap(a, b);}
    return b == 1 ? 1 : a;
}

ll lcm(ll m, ll n){ return m / gcd(m, n) * n;}

template<class T>
inline bool chmax(T &a, T b) {
    if(a < b) {a = b; return true;}
    return false;
}

template<class T>
inline bool chmin(T &a, T b) {
    if(a > b){a = b; return true;}
    return false;
}

ll power_mod(ll a, ll power, ll mod){
    ll value = 1;
    while(power != 0){
        if(power & 1) value = (value * a) % mod;
        a = (a * a) % mod;
        power = power >> 1;
    }
    return value % mod;
}

ll modinv(ll a, ll mod){
    return power_mod(a, mod - 2, mod);
}

ll power_normal(ll a, ll power){
    ll value = 1;
    while(power != 0){
        if(power & 1) value = value * a;
        a = a * a;
        power = power >> 1;
    }
    return value;
}

ll comb_memo[55][55];

ll pascal_triangle(ll n){
    comb_memo[0][0] = 1;
    For(i, 1, n + 1) rep(j, i + 1){
        comb_memo[i][j] += comb_memo[i - 1][j];
        if (j > 0) comb_memo[i][j] += comb_memo[i - 1][j - 1];
    }
}

ll combination(ll n, ll r, ll mod){
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll numerator = 1;
    ll denomenator = 1;
    for(ll i = 0; i < r ; i++){
        ll num = (n - i) % mod, den = (i + 1) % mod;
        (numerator *= num) %= mod;
        (denomenator *= modinv(den, mod)) %= mod;
    }

    return (numerator * denomenator) % mod;
}

ll combination_memo(ll n, ll r, ll pre, ll mod){
    // pre = nCr-1
    // return nCr
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll numerator = ll(n - r + 1) % mod;
    ll denomenator = modinv(r, mod) % mod;
    ll val = (numerator * denomenator) % mod;
    val *= pre;
    return val % mod;
}

ll combination_no_mod(ll n, ll r){
    if (n == r && n == 0) return 1;
    else if (n <= 0 || r < 0 || r > n) return 0;
    ll numerator = 1;
    ll denomenator = 1;
    for(ll i = 0; i < r ; i++){
        numerator *= n - i;
        denomenator *= i + 1;
        ll g = gcd(numerator, denomenator);
        numerator /= g; denomenator /= g;
    }

    return numerator;
}

ld log_combination(ll n, ll r){
    if (n == r && n == 0) return 0;
    else if (n <= 0 || r < 0 || r > n) return -INF;
    ld val = 0;
    for(ll i = 0; i < r ; i++){
        val += log(n - i);
        val -= log(i + 1);
    }

    return val;
}

ll bin_search(ll key, ll A[], ll left, ll right){
    // return the index idx where A[idx] = key.
    // A[left] is start and A[right] is end..
    // In other words, A[right], not A[right - 1], must be defined.
    while(right >= left){
        ll mid = left + (right - left) / 2;
        if (A[mid] == key) return mid;
        else if (A[mid] > key) right = mid - 1;
        else if (A[mid] < key) left = mid + 1;
    }
    return -1;
}

/*
ll bin_search_temp(ll left, ll right, callable judge){
    while(right > left){
        // when seeking lower bound
        ll mid = (right + left) / 2;
        if (judge(mid)) right = mid;
        else left = mid + 1;
        // when seeking upper bound
        ll mid = (right + left + 1) / 2;
        if (judge(mid)) left = mid;
        else right = mid - 1;
    }

    return right;
}

trinary_search

// Care the value EPS!!! Compare to the condition
ld left = 0; ld right = p;
while(abs(right - left) > EPS){
    ld left2 = (2 * left + right) / 3.0;
    ld right2 = (left + 2 * right) / 3.0;
    ld f1 = tak_func(left2);
    ld f2 = tak_func(right2);
    if (f1 <= f2) right = right2;
    else if (f2 <= f1) left = left2;
}

*/

str bin_expression(ll n, ll dig){
    str s = "";
    rep(i, dig){
        s += ltos(n % 2);
        n /= 2;
    }
    reverse(all(s));
    return s;
}

ll lower_bound_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value right instead of -1.
    if (A[left] >= key) return left;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == left) {
            if (A[left] < key && key <= A[left + 1]) return left + 1;
            else return -1;
        }
        if (A[mid - 1] < key && key <= A[mid]) return mid;
        else if (A[mid - 1] >= key) right = mid - 1;
        else if (A[mid] < key) left = mid;
    }

    return -1; // all the elements < key
}

ll inf_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value right instead of -1.
    if (A[left] > key) return left;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == left) {
            if (A[left] <= key && key < A[left + 1]) return left + 1;
            else return -1;
        }
        if (A[mid - 1] <= key && key < A[mid]) return mid;
        else if (A[mid - 1] > key) right = mid - 1;
        else if (A[mid] <= key) left = mid;
    }

    return -1; // all the elements <= key
}

ll upper_bound_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value left instead of -1.
    if (A[right] <= key) return right;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == right) {
            if (A[right - 1] <= key && key < A[right]) return right - 1;
            else return -1;
        }
        if (A[mid] <= key && key < A[mid + 1]) return mid;
        else if (A[mid] > key) right = mid;
        else if (A[mid + 1] <= key) left = mid + 1;
    }

    return -1; // all the elements > key
}

ll sup_bin_search_temp(ll key, ll A[], ll left, ll right){
    // Pay attention to the return value -1.
    // For example, sometimes you may want the value left instead of -1.
    if (A[right] < key) return right;
    ll mid = left + (right - left) / 2;

    while(right >= left){
        mid = left + (right - left) / 2;
        if (mid == right) {
            if (A[right - 1] < key && key <= A[right]) return right - 1;
            else return -1;
        }
        if (A[mid] < key && key <= A[mid + 1]) return mid;
        else if (A[mid] >= key) right = mid;
        else if (A[mid + 1] < key) left = mid + 1;
    }

    return -1; // all the elements >= key
}

ll bin_search_vector(ll key, vl v, ll left, ll right){
    // return the index idx where v[idx] = key.
    // v[left] is start and v[right] is end..
    // In other words, v[right], not v[right - 1], must be defined.
    while(right >= left){
        ll mid = left + (right - left) / 2;
        if (v[mid] == key) return mid;
        else if (v[mid] > key) right = mid - 1;
        else if (v[mid] < key) left = mid + 1;
    }
    return -1;
}

ll lower_bound_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N - 1] < key <= v[N]
    // N == -1 if all the elements < key
    return lower_bound(all(v), key) - v.begin();
}

ll inf_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N - 1] <= key < v[N] <= key + 1
    // N == -1 if all the elements <= key
    return lower_bound(all(v), key + 1) - v.begin();
}

ll upper_bound_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N] <= key < v[N + 1]
    // N == -1 if all the elements > key
    return upper_bound(all(v), key) - v.begin(); // (- 1)
}

ll sup_bin_search_vector(ll key, vl v){
    // the return value N satisfies
    // v[N] <= key - 1 < key <= v[N + 1]
    // N == -1 if all the elements >= key
    return upper_bound(all(v), key - 1) - v.begin() - 1;
}

ll fact(ll n){
    if (n == 0) return 1;
    return n * fact(n - 1);
}

ll fact_mod(ll n, ll mod){
    if (n == 0) return 1;
    return n * fact_mod(n - 1, mod);
}

bool is_prime(ll n){
    if (n <= 1) return false;
    for(ll i = 2; i * i <= n; i++){
        if (n % i == 0) return false;
    }
    return true;
}

ll bool_sum(ll a1, ll a2){
    if (a1 == 1 || a2 == 1) return 1;
    return 0;
}

mll prime_factorization(ll n){
    ll i = 2;
    mll table;
    while(i * i <= n){
        while(n % i == 0){
            table[i] ++;
            n /= i;
        }
        i++;
    }
    if (n > 1) table[n] = 1;
    return table;
}

vl divisor_table(ll n){
    vl table;
    ll i = 1;
    while(i * i <= n){
        if (n % i == 0){
            table.pb(i);
            if (i * i != n) table.pb(n / i);
        }
        i++;
    }

    sort(all(table));

    return table;
}

ll char_to_idx(char c){
    ll idx = 0;
    Forchar(cc, 'a', 'z'){
        if (c == cc) return idx;
        else idx ++;
    }
}

ll next_combination(ll sub){
    /*
    ### Attention ###
    if k is 0 or 1 and n is 0, it does not work well.

    ll n, k; ll bit = (1 << k) - 1;
    for (; bit < (1 << n); bit = next_combination(bit)){
        bool ith = bit & (1 << i);
        procedures...
    }

        sub & -sub: the binary which shares the last digit whose value is 1 in sub
        sub + x   : carry up the last digit
            ~y    : the binary whose digits are 1 if y's digit is 0.
    (sub & ~y) / x: reduce the same number of 0s after first 1 in x from (sub & ~y).
    */
    ll x = sub & -sub, y = sub + x;
    prl3(bin_expression(x, 6), bin_expression((sub & ~y), 6), bin_expression((sub & ~y) / x, 6));
    return (((sub & ~y) / x) >> 1) | y; 
}

vl z_algorithm(str s){
    ll n = s.length(); vl res(n);
    res[0] = n;
    ll i1 = 1, i2 = 0;

    while(i1 < n){
        while (i1 + i2 < n && s[i2] == s[i1 + i2]) ++i2;
        res[i1] = i2; if (i2 == 0){++i1; continue;}
        ll i3 = 1;
        while (i1 + i3 < n && i3 + res[i3] < i2){res[i1 + i3] = res[i3]; ++i3;}
        i1 += i3, i2 -= i3;
    }
    return res;
}

ll big_number_mod(str s, ll mod){
    ll l = s.length(); ll idx = 0;
    ll val = 0; ll tenth = 1;
    while (idx < l){
        ll m = ctol(s[l - 1 - idx]);
        val += (m * tenth) % mod; val %= mod;
        tenth *= 10; tenth %= mod; idx++;
    }

    return val;
}

ll string_to_ll(str s){
    ll l = s.length(); ll idx = 0;
    ll val = 0; ll tenth = 1;
    while (idx < l){
        ll m = ctol(s[l - 1 - idx]);
        val += (m * tenth);
        tenth *= 10; idx++;
    }

    return val;
}

str reflected_string(str s){
    str t, u; ll n = s.length();
    t = s; reverse(all(t));
    u = substring(t, 0, n - 2) + s + substring(t, 1, n - 1);
    return u;
}

ld distance_between_point_line(point l_begin, point l_end, point p){
    ll xl1 = l_begin.x, yl1 = l_begin.y; 
    ll xl2 = l_end.x, yl2 = l_end.y;
    ll xp = p.x, yp = p.y;
    ll a = yl2 - yl1;
    ll b = - xl2 + xl1;
    ll c = - a * xl2 - b * yl2;
    return abs(ld(a * xp + b * yp + c)) / ld(sqrt(a * a + b * b));
}

bool is_cross(point l1_begin, point l1_end, point l2_begin, point l2_end){
    ll x1 = l1_begin.x, y1 = l1_begin.y;
    ll x2 = l1_end.x, y2 = l1_end.y;
    ll x3 = l2_begin.x, y3 = l2_begin.y;
    ll x4 = l2_end.x, y4 = l2_end.y;
    ll val1 = (x1 - x2) * (y3 - y1) + (y1 - y2) * (x1 - x3);
    ll val2 = (x1 - x2) * (y4 - y1) + (y1 - y2) * (x1 - x4);
    ll val3 = (x3 - x4) * (y1 - y3) + (y3 - y4) * (x3 - x1);
    ll val4 = (x3 - x4) * (y2 - y3) + (y3 - y4) * (x3 - x2);

    return val1 * val2 < 0 && val3 * val4 < 0;
}

ld space_of_triangle(point p1, point p2, point p3){
    ll x1 = p1.x, y1 = p1.y;
    ll x2 = p2.x, y2 = p2.y;
    ll x3 = p3.x, y3 = p3.y;
    ll v1 = x2 - x1; ll u1 = y2 - y1;
    ll v2 = x3 - x1; ll u2 = y3 - y1;
    ld s = ld(v1 * u2 - u1 * v2) / ld(2);
    return abs(s);
}

pair<point, ll> center_2det_of_3points(point p1, point p2, point p3){
    ll x1 = p1.x, y1 = p1.y;
    ll x2 = p2.x, y2 = p2.y;
    ll x3 = p3.x, y3 = p3.y;
    ll c2x_2 = x2 + x1, c2y_2 = y2 + y1;
    ll c3x_2 = x3 + x1, c3y_2 = y3 + y1;
    ll b2y = x2 - x1, b2x = - y2 + y1;
    ll b3y = x3 - x1, b3x = - y3 + y1;
    ll x1_2 = c3x_2 * b3y - c3y_2 * b3x, y1_2 = c2x_2 * b2y - c2y_2 * b2x;
    ll det = - b3y * b2x + b3x * b2y;
    if (det == 0) return {{INF, INF}, det};
    ll cx_2det = - b2x * x1_2 + b3x * y1_2, cy_2det = - b2y * x1_2 + b3y * y1_2;
    return {{cx_2det, cy_2det}, det};
}

ll inversion_number(vl a, ll a_max){
    /*
    Paramters
    ---------
    a: vector<ll>
        All the elements must be non-negative.
        Prefably the elements are compressed to reduce the computational cost.
    a_max: ll
        The maximum value of the vector a or the value bigger than the value stated previously.
    */

    BIT bit(a_max + 1); ll val = 0;
    rep(i, a.size()){
        // i is the number of elements that have lower index than a[i].
        // call the number of elements that have lower value than a[i]
        // by subtracting these two, the residual number is the number of elements that have larger value 
        val += i - bit.query(a[i] - 1);// cumulative sum from 0 to a[i] - 1
        bit.update(a[i], 1);
    }
    return val;
}

template<typename T>
vector<T> compress(vector<T> v){
    // sort and remove all the duplicated values
    sort(all(v));
    v.erase(unique(all(v)), v.end());
    return v;
}

template<typename T>
map<T, ll> dict(const vector<T> &v){
    map<T, ll> d;
    rep(i, v.size()) d[v[i]] = i;
    return d;
}

/*
const ll N_VERTEX = 310;
ll a, b, t;
ll dist[N_VERTEX][N_VERTEX];

void warshall_floyd(ll n){
    // rep(i, n) rep(j, n) dist[i][j] = INF * (i != j);
    rep(k, n) rep(i, n) rep(j, n) dist[i][j] = min(dist[i][j], dist[i][k] + dist[k][j]);
}

int main(void){
    in2(n, m);
    rep(i, n) rep(j, n) dist[i][j] = INF * (i != j);
    rep(i, m){
        in3(a, b, t);
        a--; b--;
        dist[a][b] = t;
        dist[b][a] = t;
    }

    warshall_floyd(n);
}

ll dist[1005], vertex_pre[1005];

void dijkstra(ll start, ll n) {
    priority_queue<pl, vector<pl>, greater<pl>> edge_costs;
    fill(dist, dist + n, INF); // fill(vertex_pre, vertex_pre + n, -1);
    dist[start] = 0;
    edge_costs.push(pl(0, start));

    while (!edge_costs.empty()) {
        pl edge_cost = edge_costs.top();
        edge_costs.pop();
        ll idx = edge_cost.second;
        ll cost = edge_cost.first;
        if (dist[idx] < cost) continue;

        rep(i, graph[idx].size()){
            edge e = graph[idx][i];
            if (dist[e.to] > dist[idx] + e.cost){
                dist[e.to] = dist[idx] + e.cost;
                // vertex_pre[e.to] = idx;
                edge_costs.push(pl(dist[e.to], e.to));
            }
        }
    }
}

vl get_predecessor(ll g){
    vl path;
    for (; g != -1; g = vertex_pre[g]) path.pb(g);
    
    reverse(all(path));
    return path;
}

int main(void){
    in2(n, m);
    rep(i, m){
        in3(a, b, t);
        a--; b--;
        G[a].pb({b, t});
        G[b].pb({a, t});
    }

    dijkstra(0, n);
}


# ABC061D

bool find_negative_cycle(ll goal){
    rep(i, n) rep(v, n) rep(k, graph[v].size()){
        edge e = graph[v][k];
        if (dist[e.to] != INF && dist[e.to] > dist[v] + e.cost){
            dist[e.to] = -INF;
            if (goal == -1) return true;
            else if (goal == e.to) return true;
        }
    }
    return false;
}

bool bellman_ford(ll start, ll n, ll goal){
    // if there is a closed circuit, it returns false. (when goal == -1)
    // if the distance to goal cannot be obtained, it returns false (when goal != -1)
    fill(dist, dist + n, INF);
    dist[start] = 0;
    rep(i, n) rep(v, n) rep(k, graph[v].size()){
        edge e = graph[v][k];
        if (dist[v] != INF && dist[e.to] > dist[v] + e.cost) 
            dist[e.to] = dist[v] + e.cost;
    }

    if (find_negative_cycle(goal)) return false;
    return true;
}

*/

/*
    # 1. The usage of map pair
    map<pl, ll> cnt;
    cnt[{i, j}] = 0;
    items(kv, cnt){
        prl2(kv.first, kv.second);
    }

    # 2. The usage of next_permutation and combination (factorial search)
    ll a[8];
    rep(i, 8) a[i] = i;
    sort(a, a + 8);
    do{
    }while(next_permutation(a, a+n));

    // here, combination
    ll n, r;
    ll i1, i2, ..., ir;
    For(i1, n - r, n) For(i2, n - r - 1, i1) ... For(ir, n - 2 * r + 1, irr){ process;}

    # 3. bit search
    ll n;
    in1(n);
    const ll base = 3;
    ll upper = power_normal(base, n);
    
    rep(i, upper){
        ll tmp = i;
        rep(j, n){
            rep(k, base) if (tmp % base == k) prl(k);
            tmp /= base;
        }
    }

    # 4. imos method
    // used when we would like to count the number which
    // shows how many times the numbers between l and r belongs to smt.
    // This method is composed of three process.
    ll n, m, s[MAX_M], l, r;
    in2(n, m);
    rep(i, m) s[i] = 0;

    // 1st step
    rep(i, n){
        in3(l, r, c);
        l--; r--; // if l starts from 1.
        s[l] += c; s[r + 1] -= c;
    }

    // 2nd step
    rep(i, m - 1) s[i + 1] += s[i];
    // 3rd step: judgement...

    #5. shakutori method (syakutori)
    // 1. strech right side while the condition is met.
    // 2. renew the answer
    // 3. increments left side
    // 4. Back to 1. (l <= r must be satisfied all the time.)

    ll l = 0; ll r = 0;
    while (l < n){
        r = max(r, l);
        if (l == r) r++;
        while(r < n && cond) r++;
        answer += r - l; l++;
    }
    prl(answer);

    #6. priority queue

    pq q;
    ll answer = 0;
    ll v;

    rep(i, n) q.push(a[i]);
    rep(i, m){
        v = q.top(); q.pop(); // get the top value and dump the value from queue
        v /= 2; q.push(v); // add the new value
    }

    while(!q.empty()){
        answer += q.top();
        q.pop();
    }

    #7. The shortest path (distance) between the k-th edge and another edge (Tree)
    ll depth[MAX_N];
    gr tree[MAX_N];

    void dfs(ll source, ll parent, ll all_cost){
        depth[source] = all_cost;
        items(e, tree[source]){
            if (e.to == parent) continue;
            dfs(e.to, source, all_cost + e.cost);
        }
    }

    ll n, k, a, b, c;

    in2(n, k);
    rep(i, n - 1){
        in3(a, b, c);
        a--; b--;
        tree[a].pb({b, c});
        tree[b].pb({a, c});
    }

    k--;
    dfs(k, -1, 0);

    #10. Visiting Subtree using recurrent function (ABC138D)

    gr tree[MAX_N];
    ll c[MAX_N];
    bool visited[MAX_N];

    void dfs(ll source, ll parent, ll val){
        visited[source] = true;
        c[source] += val;
        rep(i, tree[source].size()){
            rep(i, m){ll res = n % match[i].e1;}
            ll vertex = tree[source][i].to;
            if (vertex == parent) continue;
            dfs(vertex, source, c[source]);
        }
    }

    #11. bfs ABC146D, ABC007C
    1. first create a tree.
    2. start searching from a node.
    3. do some processes and push nodes connected with a given target node in BFS.
    4. repeat a series of procedure until queue is empty.

    queue<pl> q;

    void bfs(ll source, ll parents){
        ll n_edge = G[source].size();
        if (parents != -1) dist[source] = min(dist[source], dist[parents] + 1);
        if (visited[source]) return;
        visited[source] = true;

        rep(idx, n_edge){
            ll target = G[source][idx].to;
            if (target == parents) continue;
            q.push(mp(target, source)); 
        }
    }

    q.push(mp(sg.e1, -1));
    while(!q.empty()){
        pl source = q.front(); q.pop();
        bfs(source.e1, source.e2);
    }

    #12. grid to distance matrix (dx, dy)
    ll w, h;
    ll pos_to_idx(ll x, ll y){
        return y * w + x;
    }

    pl idx_to_pos(ll idx){
        return mp(idx % w, idx / w); 
    }

    int main(void){
        in2(h, w);
        rep(y, h){
            in1(s);
            rep(x, w){
                if (s[x] == '#') wall[x][y] = true;
                else wall[x][y] = false;
            }
        }

        rep(i1, h * w)rep(i2, h * w) dist[i1][i2] = INF * (i1 != i2);

        rep(x, w)rep(y, h){
            ll idx1 = pos_to_idx(x, y); ll idx2;
            if (wall[x][y]) continue;
            if (x != 0 && !wall[x - 1][y]){
                idx2 = pos_to_idx(x - 1, y);
                // if warshall floyd
                dist[idx1][idx2] = 1;
                // if dijkstra
                // graph[idx1].pb({idx2, 1});
            }
            if (x != w - 1 && !wall[x + 1][y]){
                idx2 = pos_to_idx(x + 1, y);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
            if (y != 0 && !wall[x][y - 1]){
                idx2 = pos_to_idx(x, y - 1);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
            if (y != h - 1 && !wall[x][y + 1]){
                idx2 = pos_to_idx(x, y + 1);
                dist[idx1][idx2] = 1;
                // graph[idx1].pb({idx2, 1});
            }
        }
    }


*/

/*
# the operators regarding bit
& (AND), | (OR), ^ (XOR)
- (REVERSE), >> (SMALLER SHIFT)
<< (BIGGER SHIFT)


     x1: 0000 0001 0010 0101 0110 0111 0111
     x2: xxxx 0001 0011 0100 0101 1000 0110
x1 & x2: 0000 0001 0010 0100 0100 0000 0110

      x: 1001 1010  1100 1011 1101  1111
x & - x: 0001 0010  0100 0001 0001  0001
    sum: 1010 1100 10000 1100 1100 10000

x << y is x * 2 ** y
x >> y is rep(i, y) x = x // 2 

####### Attention #######
1 << i and 1LL << i is different. If programs show WA, try switch to this one.
Let S be a bit sequence and i be a non-negative integer
S & (1 << i) -> if true, i in S
S | (1 << i) -> S union {i}
S & -(1 << i) -> S - {i}
__builtin_popcount(S) -> the number of elements in S
S = 0 -> S is an empty set
__builtin_popcountl(i) -> the number of 1 in binary
S = (1 << n) - 1 -> S includes all the elements up to the n-th

#Conditional Operator
condition ? true : false;

#iterator
type declaration: auto
value reference: *itr
increment: itr++
decrement: itr--
substitution of value: *itr = smt

# inclusion-exclusion principle
|U[i = 1 to n] Ai| = sum[i = 1 to n] |Ai| - sum[i < j]|Ai ^ Aj| + ... + (-1)^(n - 1) |^[i = 1 to n]Ai|
*/

const ll MAX_N = 200005;
bool okay = false;
ll answer = 0;
ll n, a, b, c, d, m, x, y, z;

void solve(){
    
    For(i, -100, 100){
        For(j, -100, 100){
            if (power_normal(i, 5) - power_normal(j, 5) == x){
                prl2(i, j); return;
            }
        }
    }

    // check negative MOD
    // check index flow
    // check overwrite of the input variables
}
 
int main(void){
    in1(x);
    // assert(n <= 400);
    solve();
 
    return 0;
}