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

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

Note

Illustration for the first sample.

题意：给出一个数a 求两个数b和c使得a，b，c为勾股数

来自百度知道的证明：证明

/* ***********************************************Author       : rycCreated Time : 2016-08-21 SundayFile Name    : E:\acm\codeforces\368C.cppLanguage     : c++Copyright 2016 ryc All Rights Reserved************************************************ */#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>#include<map>#include<stack>using namespace std;typedef long long LL;typedef pair<int,int>pii;const int maxn=1000010;int main(){    LL n;cin>>n;    if(n<3ll){        printf("-1\n");    }    else {        LL k=n/2ll;        if(n&1){            printf("%lld %lld\n",2ll*(k*k+k),2ll*(k*k+k)+1);        }        else {            printf("%lld %lld\n",k*k-1,k*k+1);        }    }    return 0;}