Hello 2018 B. Christmas Spruce

B. Christmas Spruce

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Consider a rooted tree. A rooted tree has one special vertex called the root. All edges are directed from the root. Vertex u is called
a child of vertex v and vertex v is
called a parent of vertex u if there exists a directed edge from v to u.
A vertex is called a leaf if it doesn't have children and has a parent.

Let's call a rooted tree a spruce if its every non-leaf vertex has at least 3 leaf
children. You are given a rooted tree, check whether it's a spruce.

The definition of a rooted tree can be found here.

Input

The first line contains one integer n — the number of vertices in the tree (3 ≤ n ≤ 1 000).
Each of the next n - 1 lines contains one integer pi (1 ≤ i ≤ n - 1) —
the index of the parent of the i + 1-th vertex (1 ≤ pi ≤ i).

Vertex 1 is the root. It's guaranteed that the root has at least 2 children.

Output

Print "Yes" if the tree is a spruce and "No" otherwise.

Examples

input

4
1
1
1

output

Yes

input

7
1
1
1
2
2
2

output

No

input

8
1
1
1
1
3
3
3

output

Yes

Note

The first example:



The second example:



It is not a spruce, because the non-leaf vertex 1 has only 2 leaf
children.

The third example:



要求所有的点满足其中一个要求即可。①没有子节点  ②有至少三个没有子节点的子节点。

用vector来存图，之后利用size来判断即可

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <iostream>

#include <algorithm>

#include<queue>

#include<map>

using namespace std;

vector<int>mp[1100];

int dfs(int num)

{

   if(mp[num].size()==0)

   {

       return 1;

   }

   int ans=0;

   for(int i=0;i<mp[num].size();i++)

   {

       if(mp[mp[num][i]].size()==0)

       {

           ans++;

       }

   }

   if(ans>=3)

    return 1;

   return 0;

}

int main()

{

    int n;

    cin>>n;

    for(int i=2;i<=n;i++)

    {

        int u;

        cin>>u;

        mp[u].push_back(i);

    }

    int flag=0;

    for(int i=1;i<=n;i++)

    {

        if(dfs(i)==0)

        {

            flag=1;

            break;

        }

    }

    if(flag)

    {

        cout<<"No"<<endl;

    }

    else

    {

        cout<<"Yes"<<endl;

    }

}