Problem: 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 i likes the plane with number fi, where 1 ≤ fi ≤ n and fi ≠ 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, ..., fn (1 ≤ fi ≤ n, fi ≠ i), meaning that the i-th plane likes the fi-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 4 likes plane 1, plane 1 likes plane 2 and that is a love triangle.

In second example there are no love triangles.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long inf = (long long)1e18;
const long long M = (long long)1e9 + 7;
long long mpw(long long a, long long b, long long M = inf) {
  if (b == 0) return 1;
  if (b % 2 == 1) return a * mpw((a * a) % M, b / 2, M) % M;
  return mpw((a * a) % M, b / 2, M);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, f = 0;
  cin >> n;
  vector<long long> v(n + 1);
  for (long long i = 1; i < n + 1; i++) cin >> v[i];
  for (long long i = 1; i < n + 1; i++) {
    if (v[v[v[i]]] == i) {
      f = 1;
      break;
    }
  }
  cout << (f == 1 ? "YES" : "NO");
  return 0;
}