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.Scanner;

public class B {
	static ArrayList<Integer>[] graph;
	static int[] parent;

	static int n;

	@SuppressWarnings("unchecked")
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		graph = new ArrayList[n];
		parent = new int[n];

		for (int i = 1; i < n; i++) {
			int id = in.nextInt() - 1;
			if(graph[id] == null) graph[id] = new ArrayList<Integer>();
			parent[i] = id;
			graph[id].add(i);
		}
		System.out.println(solve() ? "Yes" : "No");
		in.close();
	}

	static boolean solve() {
		int[] counter = new int[n];
		for (int i = 1; i < n; i++) {
			if (graph[i] == null) counter[parent[i]]++;
		}

		for (int i = 0; i < n; i++) {
			if (graph[i] != null && counter[i] < 3) return false;
		}
		return true;
	}
}