time limit per test1 second

memory limit per test256 megabytes

inputstandard input

outputstandard output

Vladik and Chloe decided to determine who of them is better at math. Vladik claimed that for any positive integer n he can represent fraction as a sum of three distinct positive fractions in form .

Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that . Because Chloe can’t check Vladik’s answer if the numbers are large, he asks you to print numbers not exceeding 109.

If there is no such answer, print -1.

Input

The single line contains single integer n (1 ≤ n ≤ 104).

Output

If the answer exists, print 3 distinct numbers x, y and z (1 ≤ x, y, z ≤ 109, x ≠ y, x ≠ z, y ≠ z). Otherwise print -1.

If there are multiple answers, print any of them.

Examples

input

3

output

2 7 42

input

7

output

7 8 56

【题目链接】:

【题解】



先让x=n

则转换成

1/n=1/y+1/z;

再做一下变化得到

z=n*y/(y-n)

可以看到

如果让y=n+1

那么z的分母为1

则z肯定是一个整数了

且x=n,y = n+1,z =n*(n+1)

当n大于1的时候x

#include <bits/stdc++.h>

using namespace std;

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define LL long long

#define rep1(i,a,b) for (int i = a;i <= b;i++)

#define rep2(i,a,b) for (int i = a;i >= b;i--)

#define mp make_pair

#define pb push_back

#define fi first

#define se second

#define rei(x) scanf("%d",&x)

#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;

typedef pair<LL,LL> pll;

//const int MAXN = x;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};

const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};

const double pi = acos(-1.0);

LL n,a,b,c;

int main()

{

    //freopen("F:\\rush.txt","r",stdin);

    rel(n);

    if (n==1)

    {

        puts("-1");

    }

    else

    {

        a = n,b = n+1,c = n*n+n;

        cout << a << " " << b << " " <<c<<endl;

    }

    return 0;

}