本文共 1794 字，大约阅读时间需要 5 分钟。

Pythagorean Triples

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

若 a <=2 者不存在这样的直角三角形

若 a >2 时 ： 若 a 为偶数 ： （a^2/4+1）^2-（a^2/4-1）^2=a^2; 乘积差公式 : (a^2/4+1)-(a^2/4-1)=(a^2/2)*2=a^2; 若 a 为奇数 ： ((a^2+1)/2)^2 -(（a^2-1）/2)^2=a^2; 同上 ： ((a^2+1)/2)^2 -(（a^2-1）/2)^2=（a^2）*1=a^2;

#include
   
    int main(){    long long ma,mb,mc;    scanf("%lld",&ma);    if(ma<=2)        printf("-1\n");    else    {        if(ma%2==0)        {        ma=ma*ma;        mb=ma/4-1;        mc=ma/4+1;        printf("%lld %lld\n",mb,mc);        }       else       {        ma=ma*ma;        mb=(ma-1)/2;        mc=(ma+1)/2;        printf("%lld %lld\n",mb,mc);       }    }    return 0;}
   

转载地址：https://blog.csdn.net/WYK1823376647/article/details/52274123 如侵犯您的版权，请留言回复原文章的地址，我们会给您删除此文章，给您带来不便请您谅解！