L - Love Triangle

As you could know there are no male planes nor female planes. However, each plane on Earth likes some other plane. There are n planes on Earth, numbered from 1 to n, and the plane with number ilikes the plane with number f_i, where 1 ≤ f_i ≤ nand f_i ≠ i.

We call a love triangle a situation in which plane A likes plane B, plane B likes plane C and plane C likes plane A. Find out if there is any love triangle on Earth.

Input

The first line contains a single integer n (2 ≤ n ≤ 5000) — the number of planes.

The second line contains n integers f₁, f₂, ..., f_n(1 ≤ f_i ≤ n, f_i ≠ i), meaning that the i-th plane likes the f_i-th.

Output

Output «YES» if there is a love triangle consisting of planes on Earth. Otherwise, output «NO».

You can output any letter in lower case or in upper case.

Examples

Input

5
2 4 5 1 3

Output

YES

Input

5
5 5 5 5 1

Output

NO

Note

In first example plane 2 likes plane 4, plane 4likes plane 1, plane 1 likes plane 2 and that is a love triangle.

In second example there are no love triangles.

 

Baidu translation: As you know, there is no plane for men and women. However, each plane on the planet like the other plane. N on earth planes, numbered from 1 to n, i is the number of flat planar like number fi, and wherein 1≤fi≤n fi ≠ i. We like a triangle called the B side A side, B side surface like C, C A face like the face, to see if there triangle on the planet.

 

Idea: because love is the i-th fi, so when sz [sz [sz [i]]] is equal to i, expressed triangle. Note that the i-th value is not equal to i.

 

 1 #include <cstdio>
 2 #include <fstream>
 3 #include <algorithm>
 4 #include <cmath>
 5 #include <deque>
 6 #include <vector>
 7 #include <queue>
 8 #include <string>
 9 #include <cstring>
10 #include <map>
11 #include <stack>
12 #include <set>
13 #include <sstream>
14 #include <iostream>
15 #define mod 1000000007
16 #define eps 1e-6
17 #define ll long long
18 #define INF 0x3f3f3f3f
19 using namespace std;
20 
21 int sz[5005],n;
22 int main()
23 {
24     scanf("%d",&n);
25     for(int i=1;i<=n;i++)
26     {
27         scanf("%d",&sz[i]);
28     }
29     for(int i=1;i<=n;i++)
30     {
31         if(sz[sz[sz[i]]]==i&&sz[i]!=i)
32         {
33             printf("YES\n");
34             return 0;
35         }
36     }
37     printf("NO\n");
38 }