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:
/* package whatever; // don't place package name! */
 
import java.util.*;
import java.lang.*;
import java.io.*;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.IOException;
 
/* Name of the class has to be "Main" only if the class is public. */
public class bbg
{
	public static int result;
 
public static ArrayList<Integer> [] graph;
 
public static int[]cats;
 
public static int vizitat[];
public static int x;
//public static HashMap<String, Integer> map2;
 
 
	public static void main (String[] args) throws java.lang.Exception
{
 
Scanner input=new Scanner(System.in);
 
HashMap<Integer, Integer> contor1= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> contor2= new HashMap<Integer, Integer>();


HashMap<Integer, Integer> map= new HashMap<Integer, Integer>();
HashMap<Integer, Integer> map2= new HashMap<Integer, Integer>();
HashMap<String, Integer> combinari= new HashMap<String, Integer>();
 
 
		
		
		
		int n=input.nextInt();
		
		int v[]=new int[n+1];
		
		
		for(int i=2;i<=n;i++) {v[i]=input.nextInt();
		
		map2.put(v[i],1);
		
		if(v[i]>1)
		v[v[i]]=0;
		
	//	System.out.println(i+" "+v[i]);
		
		
		
		
		}
	//	for(int i=1;i<=n;i++)
	
		
		
		for(int i=1;i<=n;i++) 
		
		if(!map.containsKey(v[i])) map.put(v[i],1);
		else map.put(v[i],1+map.get(v[i]));
		
		int y=1;
		
		for(int x:map.keySet()) if(map.get(x)<3 && x>0) {
		    
		    	
		    		
		    y=0; break;}
		

		for(int g:map2.keySet()) if(!map.containsKey(g) || (map.containsKey(g) && map.get(g)<3)) y=0;
		if(y==0) System.out.println("No");
		else System.out.println("Yes");
	
//	for(int x:map.keySet()) System.out.println(x+" "+map.get(x));
	
		
		}
		
		
		}