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 P913B {
    
    public static void main(String[] args) {
        
        Scanner sc = new Scanner(System.in);
        
        int n = Integer.parseInt(sc.nextLine());
        
        int[] tree = new int[n];
        
        tree[0] = n;
        
        for(int i = 1; i < n; i++) {
            
            tree[i] = Integer.parseInt(sc.nextLine());
            
        }
        
        if(isSpruce(tree, 1)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
        
        sc.close();
        
    }
    
    static boolean isSpruce(int[] graph, int v) {
        
        int leaves = 0;
        boolean b = true;
        
        for(int i = v-1; i < graph.length; i++) {
            
            if(graph[i] == v) {
                if(isLeaf(graph, i+1)) {
                    leaves++;
                } else {
                    b = b && isSpruce(graph, i+1);
                }
            }
            
        }
        
        return (b && (leaves >= 3));
        
    }
    
    static boolean isLeaf(int[] graph, int v) {
        
        for(int i = v-1; i <graph.length; i++) {
            if(graph[i] == v) {
                return false;
            }
        }
        
        return true;
        
    }
    
}