Codeforces 707C Pythagorean Triples（已知直角三角形一边求另两边）

C. Pythagorean Triples

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 thatn, m andk 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

Note

Illustration for the first sample.

题目大意：

    给出直角三角形一边，求另两边，如果有多组情况，输出一组即可。

解题思路：

    打表观察可以发现，若n为偶数另外两边为(n/2*n/2-1)和(n/2*n/2+1)，若n为奇数则为(n*n)/2和(n*n)/2+1。至于证明，另奇数为2*n+1，偶数为n*2带入式子就可以发现这是一个恒等式。

    不过还需要注意，当n<=2时无解，需要特判。

    在讨论区还看到利用直角三角形另一个性质解这道题的方法：

AC代码：

#include 
using namespace std;
#define LL long long

int main()
{
    LL n;
    scanf("%I64d",&n);
    if(n==1||n==2)
        puts("-1");
    else if(n%2)
        printf("%I64d %I64d\n",n*n/2,n*n/2+1);
    else
    {
        LL p=1;
        if(n<=2)
            puts("-1");
        else if(n%2)
            printf("%I64d %I64d\n",n*n/2*p,(n*n/2+1)*p);
        else printf("%I64d %I64d\n",n/2*n/2-1,n/2*n/2+1);
    }

    return 0;
}