Codeforces Round #368 (Div. 2) Pythagorean Triples

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 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

3

output

4 5

input

6

output

8 10

input

1

output

-1

input

17

output

144 145

input

67

output

2244 2245

题目大意，给出你直角三角形任意一条边，求其他两条边，所有边均为自然数；

因为如有多组情况输出任意一种即可；

则假设n为直角三角形最短边b；

b^2=c^2-a^2;

有定理：任意两平方数之差必定为奇数或者是4的倍数.

且当n<=2时，不会有直角三角形构成

则代码可写为：

#include
typedef long long int ll;
int main()
{
    ll n;
    while(~scanf("%d",&n))
    {
        if(n <= 2)
        printf("-1\n");
            
        else if(n % 2 == 1)
        {
            ll a = (n * n + 1) / 2;
            ll c = (n * n - 1) / 2;
         printf("%I64d %I64d\n",a,c);
        else
        {
            ll a = (n * n) / 4 + 1;//若n为偶数,则n的平方比为四的倍数
            ll c = (n * n) / 4 - 1;
      
printf("%I64d %I64d\n",a,c);
 } } return 0;}