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.*;

public class B913 {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		ArrayList<Integer>[] al = new ArrayList[n];
		int[] isLeaf = new int[n]; // 0 if leaf, 1 if not
		for( int i = 0; i < n; i++ ) al[i] = new ArrayList();
		int v;
		for( int i = 0; i < n-1; i++ ) {
			v = in.nextInt()-1;
			al[v].add(i+1);
			isLeaf[v] = 1;
		}
		boolean spruce = true;
		for( int i = 0; i < n; i++ ) {
			int leaves = 0;
			if( isLeaf[i] == 1 ) { // non leaf
				for( int j = 0; j < al[i].size(); j++ ) {
					if(isLeaf[al[i].get(j)] == 0) leaves++;
				}
				if( leaves++ < 3 ) {
					spruce = false; break;
				}
			}
		}
		
		if(spruce) System.out.println("Yes");
		else System.out.println("No");
	}
}