Codeforces Round #464 (Div. 2) A. Love Triangle[判断是否存在三角恋]

A. Love Triangle

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Examples

Input

Copy

5
2 4 5 1 3

Output

YES

Input

Copy

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.

 [题意]:判断是否存在A喜欢B,B喜欢C,C喜欢A.(即三角恋)

[分析]:a[a[a[i]]]==i可以判断存在三角恋

[代码]:

#include<bits/stdc++.h>

using namespace std;
const int maxn = 5000+10;

int main()
{
    int n,a[maxn],ans,j,k;
    while(cin>>n)
    {
        ans=0;
        for(int i=1;i<=n;i++)
        {
            cin>>a[i];
        }

        for(int i=1;i<=n;i++)
        {
            j=a[i];
            k=a[j];
            if(a[k]==i && i!=j && j!=k && i!=k) ans++;
        }
        printf("%s\n",ans?"YES":"NO");
    }
}

 

 

#include<bits/stdc++.h>

using namespace std;
const int maxn = 5000+10;

int main()
{
    int n,a[maxn],f,ans;
    while(cin>>n)
    {
        for(int i=1;i<=n;i++)
            cin>>a[i];
        f=0;
        for(int i=1;i<=n;i++)
        {
            if(a[a[a[i]]]==i) f=1;
        }
        printf("%s\n",f?"YES":"NO");
    }
}