Codeforces Round #368 (Div. 2) C

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 calledPythagorean 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

output

4 5

input

output

8 10

input

output

-1

input

output

144 145

input

output

2244 2245

Note

$\text{[math]}$

Illustration for the first sample.

题意：已经一个数字，求另外两个数成为勾股数，没有就输出-1

解法：1 2就不可能有的，n为奇数，b=2n^2+2n,c=2n^2+2n+1，n为偶数b=n^2-1,c=n^2+1

#include<bits/stdc++.h>
using namespace std;
string s;
int main()
{
    int m;
    long long n;
    cin>>m;
    if(m==1||m==2)
    {
        cout<<"-1"<<endl;
    }
    else
    {
        if(m&1)
        {
            n=(m-1)/2;
            cout<<2*n*n+2*n<<" "<<2*n*n+2*n+1<<endl;
        }
        else
        {
            n=(m)/2;
            long long ans1=n*n-1;
            long long ans2=n*n+1;
            cout<<ans1<<" "<<ans2<<endl;
        }
    }
    return 0;
}