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.

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

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.

3

2 7 42

7

 //直接模拟，枚举比2/n小的分数，从1/(n/2+1)开始到2/n-1/(n/2+1)结束，这样依次得到x,y,z，记住分子分母要约分要

 //防止超过1e9，判断x,y,z是否符合条件即可。

 #include<bits\stdc++.h>

 typedef long long ll;

 using namespace std;

 ll gcd(ll x,ll y)

 {

     if(y==) return x;

     return gcd(y,x%y);

 }

 int main()

 {

     int n;

     while(cin>>n)

     {

         if(n==)

         {

             cout<<"-1\n";

             continue;

         }

         ll f1,f2,maxn,minn,f3,f4,x,y,z;

         if(n&)

         {

             f1=;f2=n;

         }

         else

         {

             f1=;f2=n/;

         }

         minn=(f2+)/f1;

         maxn=f2*minn;

         int flag=;

         for(ll i=minn;i<=maxn;i++)

         {

             x=i;

             f4=i*f2;

             f3=f1*i-f2;

             ll cnt=gcd(f4,f3);

             f4/=cnt;f3/=cnt;

             y=(f4+)/f3;

             z=y*f4;

             ll tem=f3*y-f4;

             cnt=gcd(z,tem);

             z/=cnt;tem/=cnt;

             if(x==y||x==z||z==y||tem!=||x<=||y<=||z<=||z>1e9||y>1e9||z>1e9)

                 continue;

             flag=;

             break;

         }

         if(flag) cout<<x<<" "<<y<<" "<<z<<endl;

         else cout<<"-1\n";

     }

     return ;

 }

 //本题这样做就麻烦了，其实只有n=1时不能拆成三个分数相加，其余的数都可以拆成1/n,1/(n+1),1/n*(n+1);

 #include<bits\stdc++.h>

 using namespace std;

 int main()

 {

     int n;

     cin>>n;

     if(n==) cout<<"-1\n";

     else

         cout<<n<<" "<<n+<<" "<<n*(n+)<<endl;

     return ;

 }