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 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.
Sample Input
3
4 5
6
8 10
1
-1
17
144 145
67
2244 2245
Hint
Illustration for the first sample.
假设输入的n是一条直角边的长度,那么
根据平方差公式可得
那么,这个时候,我们要求解的就是a,b
要明确，我们只不过要求解一组解即可！在对n^2划分奇偶后，只要构造出整数解即可！
接下来要做的就是解方程
于是乎,我们分类讨论即可
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define ll long long int
using namespace std;
int main()
{
ll n,a,b;
ll ans1,ans2;
scanf("%lld",&n);
if(n==||n==)
{
printf("-1\n");
return ;
}
else if(n*n%==)
{
ans1=(n*n-)/;
ans2=(n*n+)/;
}
else
{
ans1=(n*n/-)/;
ans2=(n*n/+)/;
}
printf("%lld %lld\n",ans1,ans2);
return ;
}