codeforces 743 C. Vladik and fractions(构造)

C. Vladik and fractions

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard 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

题解：2/n=1/n+1/n，那我们让其中一个是n，原问题就变成了1/n=1/x+1/y,通分变形一下(x-n)(y-n)=n*n,那我们让x-n=1,y-n=n*n即可，由于x,y,z要不同，所以当n是1的时候是不行的。

#include 
using namespace std;

#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define pb push_back
#define mp make_pair
#define cl(a) memset((a),0,sizeof(a))
#ifdef HandsomeHow
#define dbg(x) cerr << #x << " = " << x << endl
#else
#define dbg(x)
#endif
typedef long long ll;
typedef unsigned long long ull;
typedef pair  pii;
const int inf=0x3f3f3f3f;
const double eps=1e-8;
const int mod=1000000007;
const double pi=acos(-1.0);
inline void gn(long long&x){
    int sg=1;char c;while(((c=getchar())<'0'||c>'9')&&c!='-');c=='-'?(sg=-1,x=0):(x=c-'0');
    while((c=getchar())>='0'&&c<='9')x=x*10+c-'0';x*=sg;}
inline void gn(int&x){long long t;gn(t);x=t;}
inline void gn(unsigned long long&x){long long t;gn(t);x=t;}
ll gcd(ll a,ll b){return a? gcd(b%a,a):b;}
ll powmod(ll a,ll x,ll mod){ll t=1ll;while(x){if(x&1)t=t*a%mod;a=a*a%mod;x>>=1;}return t;}
// (づ°ω°)づe★------------------------------------------------

int main(){
#ifdef HandsomeHow
	//freopen("data.in","r",stdin);
	//freopen("data.out","w",stdout);
#endif
	int n;
	cin>>n;
	if(n == 1) return puts("-1"),0;
	printf("%d %d %d\n",n,n+1,n*n+n);
	return 0;
}