# Codeforces Round #368 (Div. 2) C. Pythagorean triples (Right angle)

Source: Internet
Author: User
Tags printf time limit

C. Pythagorean Triples
Time limit per test1 second
Memory limit per test256 megabytes
Inputstandard input
Outputstandard output
Katya Studies in a fifth grade. Recently her class studied right triangles and the Pythagorean theorem. It appeared, that there is triples of positive integers such that's can construct a right triangle with segments of Len Gths corresponding to triple. Such triples is called Pythagorean triples.

For example, triples (3, 4, 5), (5,,) and (6, 8, ten) are Pythagorean triples.

Here Katya wondered if she can specify the length of some side of right triangle and find any Pythagorean triple Correspon Ding 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. Would 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 integers m and K (1≤m, k≤1018), such that N, M and K form a Pythagorean triple

If there is no any Pythagorean triple containing integer n, print-1 If There is 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 Illustration for the first sample.

Test instructions: give the edge length n, find the other two side length m,k, make the triangle with n,m,k as the edge length satisfies the Pythagorean theorem. For an n, find any set of solutions, if there is no solution then output-1

Ideas
N*n+m*m=k*k;
K*k-m*m=n*n;
(k+m) (k-m) =n*n;
If N*n is an even number
K+m=n*n/2
k-m=2
If N*n is an odd number
K+m=n*n
K-m=1
A set of solutions can be found by solving a two-dollar equation group at a time.

Pay attention to the n<=2 of the special sentence

Code

```
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h >
#include <queue>
using namespace std;
int main ()
{
long long int A;
scanf ("%i64d", &a);
if (a<=2)
printf (" -1\n");
else
{
if (a*a%2==0)
printf ("%i64d%i64d\n", (a*a-4)/4, (a*a-4)/4+2);
else
printf ("%i64d%i64d\n", (a*a-1)/2, (A*a-1)/2+1);
}
return 0;
}

```
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