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 demo1 {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int a[] = new int[10000]; //第i节点的父索引
        int b[] = new int[10000]; //该父索引有的子节点数目(不管符不符合条件，反正就是子节点)
        int c[] = new int[10000]; //该父索引有的子节点数目（满足云杉树条件的子节点）
        for(int i = 2; i <= n; i++)
        {
            a[i] = in.nextInt();
            b[a[i]]++;
        }
        for(int i = 2; i <= n; i++)
        {
            if(b[i] == 0)
                c[a[i]]++;
        }
        for(int i = 1; i <= n; i++)
        {
            if(c[i] != 0 && c[i] < 3)   //有子节点  但是不满足云杉树条件
            {
                System.out.println("No");
                System.exit(0);
            }
            if(b[i] != 0 && c[i] == 0)  //有子节点  但是都不符合条件  每个子节点还有子节点
            {
                System.out.println("No");
                System.exit(0);
            }
        }
        System.out.println("Yes");
    }
}