Pythagorean Triples毕达哥斯拉三角（数学思维+构造）

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

Input

3

Output

4 5

Input

6

Output

8 10

Input

1

Output

-1

Input

17

Output

144 145

Input

67

Output

2244 2245

Hint

Illustration for the first sample.

  题目意思：给你一条边，求出另外的两条边，使得这三条边能够构造出一个直角三角形。

 

 解题思路：首先要明确一点，题目说过如果有多个解，只要输出一组就可以了，我认为这个要求很关键，只要构造出一组解就行，实际上这样也解放了思维。比如所给的边，可以是直角边，也可以是斜边，但我们知道如果是直角边那么一定可以找出一组边与其构成直角三角形；但是如果是斜边的话，则不一定能够找出一组边，所以假定所给的边为直角边更好。那么接下来分析：

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

 1 #include<iostream>
 2 #include<algorithm>
 3 #include<cstdio>
 4 #include<cstring>
 5 #define ll long long int
 6 using namespace std;
 7 int main()
 8 {
 9     ll n,a,b;
10     ll ans1,ans2;
11     scanf("%lld",&n);
12     if(n==1||n==2)
13     {
14         printf("-1\n");
15         return 0;
16     }
17     else if(n*n%2==1)
18     {
19         ans1=(n*n-1)/2;
20         ans2=(n*n+1)/2;
21     }
22     else
23     {
24         ans1=(n*n/2-2)/2;
25         ans2=(n*n/2+2)/2;
26     }
27     printf("%lld %lld\n",ans1,ans2);
28     return 0;
29 }