Codeforces Round #368 (Div. 2)C 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

勾股数有两两种经典方式（百度百科）：
1 当a为大于1的奇数2n+1时，b=2n^2+2n, c=2n^2+2n+1。
2 当a为大于4的偶数2n时，b=n^2-1, c=n^2+1
然后注意下小于3没法形成勾股数

#include<bits/stdc++.h>using namespace std;typedef long long LL;#define INF 0x3f3f3f3fconst int N=100005;int main(){    LL n;    scanf("%I64d",&n);    if(n<=2){        cout<<-1<<endl;    }    else {        LL x,y;        if(n%2){            n=(n-1)/2;            x=2*n*n+2*n;            y=x+1;        }        else {            n/=2;            x=n*n-1;            y=n*n+1;        }        cout<<x<<" "<<y<<endl;    }    return 0;}