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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader bi = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(bi.readLine());
        int[] mas = new int[n];
        int[] ans = new int[n];
        Map<Integer,Integer> map= new TreeMap<>();
        int a;
        for (int i = 1; i < n; i++) {
            a = Integer.parseInt(bi.readLine());

            mas[i] = a;
            if(map.containsKey(a))
                map.put(a,map.get(a)+1);
            else
                map.put(a,1);
        }
        for (int i = mas.length - 1; i > 0; i--) {
            if (mas[i] > 0) {
                a=mas[mas[i] - 1];
                mas[mas[i] - 1] = 0;
                if(map.containsKey(a))
                map.put(a, map.get(a) - 1);
            }
        }
        for (int i = 1; i < mas.length; i++) {
            ans[mas[i]]++;

        }
        a=0;
        for(Map.Entry<Integer, Integer> key : map.entrySet()){
            if(key.getValue()<3){
                a=1;
                System.out.println("No");
                break;
            }
        }
        if(a==0){
            System.out.println("Yes");
        }
    }
}