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

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 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 thatn, m andk 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

大体题意：
给你一个直角三角形的边n，求出另外两个边的大小（随意输出一个），不存在输出-1！
思路：
没有做出来，当时是瞎暴的，枚举角度 不是超时 就是WA ！！
赛后借鉴了大神的代码！
只能当作公式来记了！
直角三角形竟然仅仅没有1 和2 其余的都存在！
如果n能整除4，那就是a/4*3和 a/4*5了！
如果n能整除2  那就是(a/2)*(a/2)/2 * 2,和 （(a/2)*(a/2)/2 +1）* 2
否则对于一般的数那就是a*a/2和 a*a/2+1！

#include<bits/stdc++.h>using namespace std;typedef long long ll;int main(){    ll a;    scanf("%I64d",&a);    if (a == 1 || a == 2){        printf("-1\n");        return 0;    }    if (a % 4 == 0){        printf("%I64d %I64d\n",a/4*3,a/4*5);        return 0;    }    if (a % 2 == 0){        ll t = a/2;        printf("%I64d %I64d\n",(t*t/2)*2, (t*t/2+1)*2 );        return 0;    }    printf("%I64d %I64d\n",a*a/2,a*a/2+1);    return 0;}