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.Scanner;

public class ChristmasSpruce {

	/**
	 * @param args
	 */
	public static void main(String[] args){
		// TODO Auto-generated method stub

		
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int p[] = new int[n+1];
		int c[] = new int[n+1];
		int l[] = new int[n+1];
		c[1] = 1;
		for(int i=2;i<n+1;i++){
			p[i] = sc.nextInt();
			c[p[i]] = 1;
		}
		for(int i=2;i<=n;i++){
			if(c[i] == 0)
				l[p[i]]++;
		}
		String str = "Yes";
		for(int i=1;i<=n;i++){
			if(c[i] == 1 && l[i]<3){
				str = "No";
				break;
			}
		}
		System.out.println(str);
	}
}