Given an integer x, find 2 integers a and b such that: 

  * 1 ≤ a,b ≤ x 
  * b divides a (a is divisible by b). 
  * a ⋅ b>x. 
  * a/b<x. 

Input

The only line contains the integer x (1 ≤ x ≤ 100).

Output

You should output two integers a and b, satisfying the given conditions, separated by a space. If no pair of integers satisfy the conditions above, print "-1" (without quotes).

Examples

Input

10


Output

6 3

Input

1


Output

-1
#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1.0);
const int mod = (1 ? 1000000007 : 998244353);
const int inf = (1 ? 0x3f3f3f3f : 0x7fffffff);
const int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
const long long INF = (1 ? 0x3f3f3f3f3f3f3f3f : 0x7fffffffffffffff);
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long poww(long long a, long long b) {
  long long s = 1;
  while (b) {
    if (b & 1) s = (s * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return s % mod;
}
struct quickread {
  template <typename _Tp>
  inline quickread& operator>>(_Tp& x) {
    x = 0;
    char ch = 1;
    int fh = 1;
    while (ch != '-' && (ch > '9' || ch < '0')) ch = getchar();
    if (ch == '-') {
      fh = -1, ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = (x << 1) + (x << 3) + (ch ^ 48);
      ch = getchar();
    }
    x *= fh;
    return *this;
  }
} cinn;
const int N = 2e5 + 11;
long long n, m;
void solve() {
  cin >> n;
  if (n == 1)
    cout << -1 << "\n";
  else
    cout << n << ' ' << n << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long T = 1;
  while (T--) solve();
  return 0;
}
