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 code913B {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int len = sc.nextInt();
		boolean[] notLeaf = new boolean[len+1];
		boolean[] qwq = new boolean[len+1];
		int[] qaq = new int[len+1];
		int[] root = new int[len+1];
		for(int i=2;i<len+1;i++){
			int c = sc.nextInt();
			root[i] = c;
			qaq[c] ++;
			notLeaf[c] = true;
			if(!qwq[c]){
				qaq[root[c]]--;
				qwq[c] = true;
			}
		}
		for(int i=1;i<len+1;i++){
			if(notLeaf[i] & qaq[i]<3){
				System.out.println("No");
				return;
			}
		}
		System.out.println("Yes");
	}
}