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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.

百度翻译：正如你所知道的，没有男性和女性的平面。然而，地球上的每一个平面都喜欢另一个平面。地球上有n个平面，编号从1到n，而编号为i的平面喜欢编号为fi的平面，其中1≤fi≤n和fi≠i。我们把一个三角叫做A面喜欢B面，B面喜欢C面，C面喜欢A面，看看地球上有没有三角。

思路：因为是第i个喜欢fi，所以当sz[sz[sz[i]]]等于i时，表示有三角关系。注意第i个的值不能等于i.

#include <cstdio>
#include <fstream>
#include <algorithm>
#include <cmath>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <cstring>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <iostream>
#define mod 1000000007
#define eps 1e-6
#define ll long long
#define INF 0x3f3f3f3f
using namespace std;

int sz[5005],n;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&sz[i]);
    }
    for(int i=1;i<=n;i++)
    {
        if(sz[sz[sz[i]]]==i&&sz[i]!=i)
        {
            printf("YES
");
            return 0;
        }
    }
    printf("NO
");
}