Codeforces Round #368 (Div. 2) Pythagorean Triples_codeforces毕达哥拉斯三重奏-优快云博客

本文链接：https://blog.youkuaiyun.com/u013555159/article/details/52264296

本文介绍了一个算法，用于根据给定的正整数找到一组勾股数。通过判断给定数值为奇数或偶数，算法能够计算出另外两个勾股数，使三者构成直角三角形的边长。

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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

就按照套路完成就可以了，注意bc要用long long 来完成
//
//  main.cpp
//  pc
//
//  Created by 张嘉韬 on 16/8/20.
//  Copyright © 2016年 张嘉韬. All rights reserved.
//

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
int main(int argc, const char * argv[]) {
    int  a;
    scanf("%d",&a);
    if(a<=2) printf("-1\n");
    else
    {
        long long n,b,c;
        if(a%2==1)
        {
             n=(a-1)/2;
             b=2*n*n+2*n;
             c=b+1;
        }
        else
        {
             n=a/2;
             b=n*n-1;
             c=n*n+1;
        }
        //printf("%lld %lld\n",b,c);
        cout<<b<<" "<<c<<endl;
    }
    return 0;
}