CF2.C

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 $\text{[math]}$ as a sum of three distinct positive fractions in form $\text{[math]}$ .

Help Vladik with that, i.e for a given n find three distinct positive integers x, y and z such that $\text{[math]}$ . 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

Output

2 7 42

Input

Output

7 8 56

题意：

给出n，问有没有三个数x,y,z,使得1/x+1/y+1/z=2/n;

代码：

 1 //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
 2 //防止超过1e9，判断x,y,z是否符合条件即可。
 3 #include<bitsstdc++.h>
 4 typedef long long ll;
 5 using namespace std;
 6 ll gcd(ll x,ll y)
 7 {
 8     if(y==0) return x;
 9     return gcd(y,x%y);
10 }
11 int main()
12 {
13     int n;
14     while(cin>>n)
15     {
16         if(n==1)
17         {
18             cout<<"-1
";
19             continue;
20         }
21         ll f1,f2,maxn,minn,f3,f4,x,y,z;
22         if(n&1)
23         {
24             f1=2;f2=n;
25         }
26         else
27         {
28             f1=1;f2=n/2;
29         }
30         minn=(f2+1)/f1;
31         maxn=f2*minn;
32         int flag=0;
33         for(ll i=minn;i<=maxn;i++)
34         {
35             x=i;
36             f4=i*f2;
37             f3=f1*i-f2;
38             ll cnt=gcd(f4,f3);
39             f4/=cnt;f3/=cnt;
40             y=(f4+1)/f3;
41             z=y*f4;
42             ll tem=f3*y-f4;
43             cnt=gcd(z,tem);
44             z/=cnt;tem/=cnt;
45             if(x==y||x==z||z==y||tem!=1||x<=0||y<=0||z<=0||z>1e9||y>1e9||z>1e9)
46                 continue;
47             flag=1;
48             break;
49         }
50         if(flag) cout<<x<<" "<<y<<" "<<z<<endl;
51         else cout<<"-1
";
52     }
53     return 0;
54 }
55 //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);
56 #include<bitsstdc++.h>
57 using namespace std;
58 int main()
59 {
60     int n;
61     cin>>n;
62     if(n==1) cout<<"-1
";
63     else
64         cout<<n<<" "<<n+1<<" "<<n*(n+1)<<endl;
65     return 0;
66 }