Codeforces Round #368 (Div. 2)-C. Pythagorean Triples

原题链接

C. Pythagorean Triples

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

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

直角三角形a, b, c可表示为

a = m^2 - n^2;

b = 2*n*m;

c = n^2 + m ^ 2;

对于题中给出的k，若k小于等于2则无解，若k为偶数,则k = 2 * 1 * (k / 2),若k为奇数则 k = (k/2+1)*(k/2+1)-(k/2)*(k/2)

#include <iostream>#include <algorithm>#include <cstring>#include <cstdio>#include <vector>#include <map>#include <cmath>#include <queue>#define maxn 100005#define INF 1e18using namespace std;typedef long long ll;int main(){ll n;scanf("%I64d", &n);if(n <= 2){puts("-1");return 0;}if(n&1){printf("%I64d %I64d\n", 2*(n/2+1)*(n/2), (n/2+1)*(n/2+1)+(n/2)*(n/2));}else{    printf("%I64d %I64d\n", (n/2)*(n/2)-1, (n/2)*(n/2)+1);}return 0;}