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.*;
import java.io.*;

public class CF913B {
	static List<Integer>[] adj;
	static boolean spruce;
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int n = scan.nextInt();
		adj = new ArrayList[n];
		for(int i = 0 ; i < n ; i++)
			adj[i] = new ArrayList<>();
		for(int i = 0 ; i < n-1 ; i++) {
			int v = i+1;
			int u = scan.nextInt()-1;
			adj[u].add(v); // edge from parent to child
		}
		spruce = true;
		dfs(0);
		System.out.println(spruce ? "Yes" : "No");
	}
	static int dfs(int u) {
		int c = 0, lc = 0;
		for(int v : adj[u]) {
			c++;
			lc += (dfs(v) == 0 ? 1 : 0);
		}
		if(c > 0 && lc < 3)
			spruce = false;
		return c;
	}
}