Problem: 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](https://goo.gl/1dqvzz).

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:

<image>

The second example:

<image>

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

The third example:

<image>

Solution (in JAVA):

Solution:

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

/*
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/**
 *
 * @author rn-sshawish
 */
public class Contest913B {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int n = input.nextInt();
        Tree tree = new Tree(n);
        for (int i = 2; i <= n; i++) {
            tree.add(input.nextInt(), i);
        }
        System.out.println(tree.bfs(1)?"Yes":"No");
    }
}


class Tree {
    Node [] nodes;

    public Tree(int n) {
        nodes = new Node[n+1];
        init();
    }
    
    public void init(){
        for (int i = 1; i < nodes.length; i++) {
            nodes[i] = new Node(i);
        }
    }
    
    public void add(int index,int edge){
        nodes[index].edge.add(edge);
    }
    
    public boolean bfs (int root){
        
        Queue<Integer> data = new LinkedList<>();
        data.add(root);
        
        while (!data.isEmpty()) {
            
            int x  =  data.poll();
            
            int size = nodes[x].edge.size();
            
            for (Integer object : nodes[x].edge) {
                if (!nodes[object].edge.isEmpty()) {
                    size--;
                }
                data.add(object);
            }
            if (size < 3 && nodes[x].edge.size()!= 0) {
                return false;
            }
        }
        
    
        return true;
    }
}


class Node {
    int value;
    ArrayList<Integer> edge;

    public Node(int value) {
        this.value = value;
        this.edge = new ArrayList<>();
    }
    
}