codeforces 707C 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 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.

输入n,输出m, k;  使得(n, m, k)是一组勾股数。如果有多种情况输出其中一种就好

(1)n<=2时，不可能

(2)n为奇数时，m=(n*n-1)/2, k=(n*n-1)/2+1;

(3)n为偶数时，将其化为奇数来做，就是连续除以二化为奇数

1)当n为4的倍数时连续除以2最终会化为1，所以这种情况先直接对(3, 4, 5)对应成比例即可

2)当n不为4的倍数时，连续除以2最终会化为大于1的奇数，然后按(2)来做

AC代码：

# include<iostream>using namespace std;int main(){unsigned long long n;unsigned long long d=1;cin>>n;if(n<=2){cout<<-1;return 0;}if (n%4==0){cout<<(long long)3*n/4<<" "<<(long long)5*n/4<<endl;return 0;}while(n%2==0) {d*=2;n/=2;}unsigned long long now=(n*n-1)/2;if (now==1){cout<<-1<<endl;return 0;}cout<<d*now<<" "<<d*(now+1)<<endl;return 0;}