A. Ehab and another construction problem

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Given an integer xx, find 2 integers aa and bb such that:

  • 1≤a,b≤x1≤a,b≤x
  • bb divides aa (aa is divisible by bb).
  • a⋅b>xa⋅b>x.
  • ab<xab<x.

Input

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

Output

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

Examples

input

Copy


10


output

Copy


6 3


input

Copy


1


output

Copy


-1


 

#include<cstdio>
#include<iostream>
#include<fstream>
#include<algorithm>
#include<functional>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iomanip>
#include<numeric>
#include<cctype>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<list>
#include<set>
#include<map>
using namespace std;
#define N 100000+5
#define rep(i,n) for(int i=0;i<n;i++)
#define sd(n) scanf("%d",&n)
#define sll(n) scanf("%lld",&n)
#define pd(n) scanf("%d\n",n)
#define pll(n) scanf("%lld\n",n)
#define MAX 26
typedef long long ll;
const ll mod=1e9+7;
ll n,m;
ll a[N];
ll b[N];
int main()
{
     //string s;
     //cin>>s;
    ll ans=0;
     //ll sum=0;
    
     ll x,y,d;
    scanf("%lld",&n);
    int flag=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=n;j++)
        {
         
        if(i%j==0&&i*j>n&&i/j<n)
        {
            flag=1;
            x=i;y=j;
            break;
        }
        
        }if(flag==1) break;
    }  
    if(flag==1)
    printf("%lld %lld\n",x,y);
    else
    {
     printf("-1\n");    
    }
        
    
    
    
    /*for(int i=0;i<n;i++)
    {
        scanf("%lld%lld",&a[i],&b[i]);
    
    } */
    //for(int i=1;i<=n;i++)
    //  printf("%lld",a[i]);
        
 
    return 0;

}