codeforces C. Pythagorean Triples (【数学】毕达哥拉斯三元数组勾股定理)-优快云博客

本文介绍了一个算法，用于根据给定的正整数n找到一组勾股数，即能够构成直角三角形三边长度的三个正整数。文章提供了一种有效的方法来解决这个问题，并附带了一个简单的C++代码实现。

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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

Note

$\text{[math]}$

Illustration for the first sample.

【分析】
对于直角三角形△ABC,如图:

我们知道的是

因为题目说如果题目存在多种解,输出任何一个就可以了,所以我们不妨假设输入的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

于是乎,我们分类讨论即可

【时间复杂度&&优化】
O(1)

开始想着暴力，但是数据大，不可能实现，后来看题解，原来这样^^(上面推导过程)

代码：

#include<cstdio>
#define LL __int64
int main()
{
	LL n;
	scanf("%I64d",&n);
	if(n==1||n==2)	//特殊考虑 
		printf("-1\n");
	else	//结论 
	{
		if(n*n%2==1)
			printf("%I64d %I64d\n",(n*n+1)/2,(n*n-1)/2);
		else
			printf("%I64d %I64d\n",(n*n+4)/4,(n*n-4)/4);
	}		
	return 0;
}