三角形问题

Description

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.

Sample Input

Input

3

Output

4 5

Input

6

Output

8 10

Input

1

Output

-1

Input

17

Output

144 145

Input

67

Output

2244 2245

Hint

题意：给你一个直角边  输出另一条直角边和斜边    a b c  满足a*a+b*b=c*c.

当a是奇数时  满足c-b=1 联立上式  解得b=(a*a-1)/2;   c=b+1;

当a是偶数时  满足c-b=2 解得b=(a*a-4)/4  c=b+2;

#include <algorithm>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <deque>
#include <map>
using namespace std;
typedef long long ll;
map<int,int> M;
int main()
{
    ll n,i,a,b,c;
    cin>>a;
    if(a%2!=0)
    {
        b=(a*a-1)/2;
        c=b+1;
        if(b<=0||c<=0)
            cout<<"-1"<<endl;
        else
            cout<<b<<" "<<c<<endl;
    }
    else
    {
        b=(a*a-4)/4;
        c=b+2;
        if(b<=0||c<=0)
            cout<<"-1"<<endl;
        else
            cout<<b<<" "<<c<<endl;
    }
    return 0;
}