CodeForces 939A（水题）

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Description

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 f_i, where 1 ≤ f_i ≤ n and 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.

Sample Input

Input

5
2 4 5 1 3

Output

YES

Input

5
5 5 5 5 1

Output

NO

Hint

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.

我也不知道你是怎么想的，谁说的如果从第一个数开始找找不到的话这个序列就没有三角恋了呢？会死循环？？？我打你啊！辛酸。

看一下这个题怎么做吧，这个题的特点是他是一个有限的查找，对于每一个数查三次就可以判断该数是否构成了三角恋，而不需要一开始想的那么复杂，从第一个数开始，模拟题目查找的过程。

对于这种有限次且次数不多的查找，一步到底进行判断。

AC代码：

#include<stdio.h>
#include<stdio.h>

int main()
{
    int number[5009];
    int i,p,j,y,n;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
        scanf("%d",&number[i]);
    for(i=1;i<=n;i++)
        if(number[number[number[i]]]==i)
        {
            printf("YES\n");
            break;
        }
    if(i>n)
        printf("NO\n");
    return 0;
}