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 B {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int p[] = new int[n + 1];
        int childs[] = new int[n + 1];
        for(int i = 2;i <= n;i++) {
            p[i] = scanner.nextInt();
            childs[p[i]]++;
        }
        int leavesChilds[] = new int[n + 1];
        for(int i = 2;i <= n;i++)
            if(childs[i] == 0)
                leavesChilds[p[i]]++;
        boolean spruce = true;
        for(int i = 1;i <= n;i++)
            if(childs[i] != 0 && leavesChilds[i] < 3)
            spruce = false;
        if(spruce)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}