#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
long long pwr(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long temp = pwr(a, b / 2, mod);
  temp = (temp * temp) % mod;
  if (b & 1) temp = (temp * a) % mod;
  return temp;
}
long long pwr(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = pwr(a, b / 2);
  temp = (temp * temp);
  if (b & 1) temp = (temp * a);
  return temp;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long modularInverse(long long a, long long m) { return pwr(a, m - 2, m); }
bool *isPrime;
void generatePrimeSieve(const int lim) {
  isPrime = (bool *)malloc(lim + 1);
  memset(isPrime, true, lim + 1);
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i <= lim; ++i)
    if (isPrime[i])
      for (int j = i + i; j <= lim; j += i) isPrime[j] = false;
}
vector<vector<int> > identityMatrix;
vector<vector<int> > mul(const vector<vector<int> > &a,
                         const vector<vector<int> > &b) {
  int n = a.size();
  vector<vector<int> > ans(n, vector<int>(n));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        ans[i][j] += ((long long)a[i][k] * b[k][j]) % mod;
        ans[i][j] %= mod;
      }
    }
  }
  return ans;
}
vector<vector<int> > pwr(const vector<vector<int> > &a, long long n) {
  if (n == 0) {
    assert(false);
    return identityMatrix;
  }
  if (n == 1) return a;
  vector<vector<int> > tmp = pwr(a, n / 2);
  tmp = mul(tmp, tmp);
  if (n & 1) tmp = mul(a, tmp);
  return tmp;
}
int abso(int x, int y) {
  if (x < y) return y - x;
  return x - y;
}
struct compare {
  bool operator()(const int &l, const int &r) { return l > r; }
};
long long vert[100005];
long long subsz[100005];
vector<pair<int, long long> > adj[100005];
void dfs1(int v, int prev) {
  int temp;
  subsz[v] = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    temp = adj[v][i].first;
    if (temp != prev) {
      dfs1(temp, v);
      subsz[v] += subsz[temp];
    }
  }
}
long long ans = 0;
void dfs2(int v, int prev, long long dist) {
  int temp;
  if (dist > vert[v]) {
    ans += subsz[v];
    return;
  }
  for (int i = 0; i < adj[v].size(); i++) {
    temp = adj[v][i].first;
    if (temp != prev) {
      long long val = max((long long)0, dist + adj[v][i].second);
      dfs2(temp, v, val);
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  if (n == 1 || n == 2) {
    cout << "-1" << endl;
    return 0;
  }
  if (n % 2 == 0) {
    long long temp = (n * n) / 4;
    cout << temp - (long long)1 << " " << temp + (long long)1 << endl;
  } else {
    long long temp = (n * n) / 2;
    cout << temp << " " << temp + (long long)1 << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

Katya studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there are triples of positive integers such that you can construct a right triangle with segments of lengths corresponding to triple. Such triples are called Pythagorean triples.

For example, triples (3, 4, 5), (5, 12, 13) and (6, 8, 10) are Pythagorean triples.

Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple corresponding to such length? Note that the side which length is specified can be a cathetus as well as hypotenuse.

Katya had no problems with completing this task. Will you do the same?

Input

The only line of the input contains single integer n (1 ≤ n ≤ 109) — the length of some side of a right triangle.

Output

Print two integers m and k (1 ≤ m, k ≤ 1018), such that n, m and k form a Pythagorean triple, in the only line.

In case if there is no any Pythagorean triple containing integer n, print  - 1 in the only line. If there are many answers, print any of them.

Examples

Input

3


Output

4 5

Input

6


Output

8 10

Input

1


Output

-1

Input

17


Output

144 145

Input

67


Output

2244 2245

Note

<image>

Illustration for the first sample.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
long long pwr(long long a, long long b, long long mod) {
  if (b == 0) return 1;
  long long temp = pwr(a, b / 2, mod);
  temp = (temp * temp) % mod;
  if (b & 1) temp = (temp * a) % mod;
  return temp;
}
long long pwr(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = pwr(a, b / 2);
  temp = (temp * temp);
  if (b & 1) temp = (temp * a);
  return temp;
}
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
long long modularInverse(long long a, long long m) { return pwr(a, m - 2, m); }
bool *isPrime;
void generatePrimeSieve(const int lim) {
  isPrime = (bool *)malloc(lim + 1);
  memset(isPrime, true, lim + 1);
  isPrime[0] = false;
  isPrime[1] = false;
  for (int i = 2; i <= lim; ++i)
    if (isPrime[i])
      for (int j = i + i; j <= lim; j += i) isPrime[j] = false;
}
vector<vector<int> > identityMatrix;
vector<vector<int> > mul(const vector<vector<int> > &a,
                         const vector<vector<int> > &b) {
  int n = a.size();
  vector<vector<int> > ans(n, vector<int>(n));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        ans[i][j] += ((long long)a[i][k] * b[k][j]) % mod;
        ans[i][j] %= mod;
      }
    }
  }
  return ans;
}
vector<vector<int> > pwr(const vector<vector<int> > &a, long long n) {
  if (n == 0) {
    assert(false);
    return identityMatrix;
  }
  if (n == 1) return a;
  vector<vector<int> > tmp = pwr(a, n / 2);
  tmp = mul(tmp, tmp);
  if (n & 1) tmp = mul(a, tmp);
  return tmp;
}
int abso(int x, int y) {
  if (x < y) return y - x;
  return x - y;
}
struct compare {
  bool operator()(const int &l, const int &r) { return l > r; }
};
long long vert[100005];
long long subsz[100005];
vector<pair<int, long long> > adj[100005];
void dfs1(int v, int prev) {
  int temp;
  subsz[v] = 1;
  for (int i = 0; i < adj[v].size(); i++) {
    temp = adj[v][i].first;
    if (temp != prev) {
      dfs1(temp, v);
      subsz[v] += subsz[temp];
    }
  }
}
long long ans = 0;
void dfs2(int v, int prev, long long dist) {
  int temp;
  if (dist > vert[v]) {
    ans += subsz[v];
    return;
  }
  for (int i = 0; i < adj[v].size(); i++) {
    temp = adj[v][i].first;
    if (temp != prev) {
      long long val = max((long long)0, dist + adj[v][i].second);
      dfs2(temp, v, val);
    }
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  if (n == 1 || n == 2) {
    cout << "-1" << endl;
    return 0;
  }
  if (n % 2 == 0) {
    long long temp = (n * n) / 4;
    cout << temp - (long long)1 << " " << temp + (long long)1 << endl;
  } else {
    long long temp = (n * n) / 2;
    cout << temp << " " << temp + (long long)1 << endl;
  }
  return 0;
}

```