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:
// Marcelino Galarza
// January 25th, 2018

import java.util.*;

public class ChristmasSpruce 
{

	public static void main(String args[])
	{
		Scanner in = new Scanner(System.in);
		int numVertix = in.nextInt();
		ArrayList<Integer> boss[] = new ArrayList[numVertix];
		for (int i = 0; i < numVertix; i++)
			boss[i] = new ArrayList<Integer>();
		
		for (int i = 1; i < numVertix; i++)
		{
			int man = in.nextInt()-1;
			boss[man].add(i);
		}
		
		if (BFS(boss, 0))
			System.out.println("Yes");
		else
			System.out.println("No");
	}
	
	// I think I want to look ahead from my current node. If I have at least
	// three children that are leaves (size == 0) then continue by adding the 
	// children that aren't leaves. 
	public static boolean BFS(ArrayList<Integer> boss[], int index)
	{
		ArrayDeque<Integer> q = new ArrayDeque<Integer>();
		q.add(index);
		
		while (!q.isEmpty())
		{
			int at = q.poll();
			int counter = 0;
			
			for (int i = 0; i < boss[at].size(); i++)
			{
				int child = boss[at].get(i);
				if (boss[child].size() == 0)
					counter++;
				else
					q.add(boss[at].get(i));
			}

			if (counter < 3)// && boss[at].size() >= 1)
				return false;
		}	
		return true;
	}
}

//4
//1
//1
//1

// Yes

//7
//1
//1
//1
//2
//2
//2

// No

//8
//1
//1
//1
//1
//3
//3
//3

// Yes