Codeforces Round #464 (Div. 2)A. Love Triangle

 题目

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.

反思

一开始想的是找一个三个点的环, 但不允许访问访问过的点就会判不了环

允许访问有的数据就会死循环

代码

#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
#define _for(i,a,b) for(int i=(a);i<(b);i++)
#define _rep(i,a,b) for(int i=(a);i<=(b);i++)
typedef pair<int, int> PLL;
typedef long long ll;
const int N = 5e3+5;
const int INF = 0x3f3f3f3f;
int readint(){
    int x;scanf("%d",&x);return x;
}
int n;
int f[N];
int main()
{
    n = readint();
    _rep(i, 1, n) f[i] = readint();
    bool ok = false;
    _rep(i, 1, n) 
        // 我喜欢的人喜欢的人喜欢我 -> 三角恋
        if(f[f[f[i]]] == i) ok = true;
    if(ok) puts("YES");
    else puts("NO");

    return 0;
}