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.io.*;
import java.util.StringTokenizer;
import java.util.*;

public class Solution {
    private static final long mod = 1000000007;
    static int ans = 0;
    public static void main(String args[]) throws Exception {

        FastScanner sc = new FastScanner();
        PrintWriter out=new PrintWriter(System.out);
        int t;
        t = 1;//sc.nextInt();

        while (t-- > 0) {
            solve(sc,out);
        }
        out.close();

    }

    public static void solve(FastScanner sc, PrintWriter out) {
        int n = sc.nextInt();
        ArrayList<Integer> al[] = new ArrayList[n+1];
        for(int i = 1;i<=n;i++)
            al[i] = new ArrayList<>();

        for(int i = 2;i<=n;i++){
            int p = sc.nextInt();
            al[p].add(i);
        }

        for(int i = 1;i<=n;i++) {
            int count = 0;
            if(al[i].size() != 0) {
                for (int j : al[i])
                    if (al[j].size() == 0) count++;
                if (count < 3) {
                    out.println("No");
                    return;
                }
            }
        }
        out.println("Yes");
            out.flush();

    }


    public static void dfs(int node, int parent, int[] cats, ArrayList<Integer>[] edges, int catsFound, int m){

        if( catsFound <= m) {
            if (cats[node] == 1) catsFound++;
            else
                catsFound = 0;
        }

        for(int child : edges[node]){
            if(child != parent)
                dfs(child,node,cats,edges,catsFound,m);
        }
        if(edges[node].size() == 1 && edges[node].get(0) == parent && catsFound <= m)
        {
            ans++;
        }
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        public String next() {
            while (!st.hasMoreElements())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

    }
}