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;
import java.io.*;

public class B {
	static ArrayList<Integer>[] tree;
	static boolean visited[];
	static int levels = 0;

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		tree = new ArrayList[n + 1];
		visited = new boolean[n];
		for (int i = 1; i <= n; i++)
			tree[i] = new ArrayList<Integer>();

		for (int i = 0, num; i < n - 1; i++) {
			num = sc.nextInt();
			tree[num].add(i + 2);
		}

		

		Queue<Integer> q = new LinkedList<Integer>();
		q.add(1);
		
		int countLeave = 0;

		while (!q.isEmpty()) {
			int u = q.peek();
			q.poll();
			int size = tree[u].size();

			for (int i = 0; i < size; i++) 
				if (tree[tree[u].get(i)].size() == 0) countLeave++;
				else 
					q.add(tree[u].get(i));

			if (countLeave < 3) {
				System.out.println("No");
				return;
			}
			countLeave = 0;

		}
		System.out.println("Yes");
	}
}