CodeForces - 707C

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


构造直角三角形，设已知边为x，另两条边分别为x，y。
则有a*a=y*y-x*x.即a*a=(y-x)(y+x),令A=y-x,B=y+x;
解得 x=(A+B)/2,y=(A-B)/2.

现在判断什么样的A，B可使x和y是整数。
分类讨论，当a*a是偶数的时候，A=2,B=a*a/2.
当a*a是奇数的时候，A=1,B=a*a.

若解不出来，则输出-1.

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<queue>
#include<stack>
#include<deque>
#include<map>
#include<iostream>
using namespace std;
typedef long long  LL;
const double pi=acos(-1.0);
const double e=exp(1);
const int N = 10;

int main()
{
    LL i,p,j;
    LL x,y,a;
    scanf("%lld",&a);
    if(a%2==0)
    {
        x=(a*a/2-2)/2;
        y=(a*a/2+2)/2;
        if(x<=0||y<=0||x==y)
            printf("-1\n");
        else
            printf("%lld %lld\n",x,y);
    }
    else
    {
        x=(a*a-1)/2;
        y=(a*a+1)/2;
        if(x<=0||y<=0||x==y)
            printf("-1\n");
        else
            printf("%lld %lld\n",x,y);
    }
    return 0;
}