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?
The only line of the input contains single integer n (1 ≤ n ≤ 109) — the length of some side of a right triangle.
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.
3
4 5
6
8 10
1
-1
17
144 145
67
2244 2245
![](http://codeforces.com/predownloaded/1e/c5/1ec504abd66f20b75c88d3f4f370b3724f1e4e28.png)
Illustration for the first sample.
看了好多个题解,感觉这个比较容易懂,这个题真的好锻炼思维。。。打死也想不到。。。
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; int main() { long long n; while(~scanf("%lld",&n)){ if(n<=2){ printf("-1 "); }else{ if(n*n&1){ printf("%lld %lld ",(n*n-1)/2,(n*n+1)/2); }else{ printf("%lld %lld ",n*n/4-1,n*n/4+1); } } } return 0; }