codeforces.com/707/C C. Pythagorean Triples 勾股数

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 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

Note

Illustration for the first sample.

又是个数学问题，我也是醉了

这个问题是关于勾股数的，知一求二

当 a 为大于 1 的奇数 2 * n+1 时, b = 2 * n * n + 2 * n, c = 2 * n * n + 2 * n + 1.

　　当 a 为大于 4 的偶数 2 * n 时，b = n * n - 1, c = n * n + 1.

然后不满足上述构造方法的数 1, 2, 4直接特判就好.

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<cmath>#include<string>#include<vector>#include<stack>#include<set>#include<map>#include<queue>#include<algorithm>using namespace std;int main(){long long a;scanf("%lld",&a);long long b,c;long long n;if(a%2==1&&a>1){n=(a-1)/2;b=n*n*2+2*n;c=b+1;}else if(a%2==0&&a>4){n=a/2;b=n*n-1;c=b+2;}else{if(a==4){b=3;c=5;}}if(a==1||a==2){printf("-1\n");}else{printf("%lld %lld\n",b,c);}    return 0;}