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 PYTHON):

Solution:
from fractions import gcd
from math import *
from itertools import *
import string
import copy
import random
import bisect
from decimal import *
def id_generator(size=20, chars=string.digits):
    return ''.join(random.choice(chars) for _ in range(size))
 
def mp():
    return map(int,str(raw_input()).split())
 
n=input()
flag=1
l=[[] for i in range(n+1)]
for i in range(n-1):
	s=input()
	l[s].append(i+2)

for i in range(1,n):
	val=0
	for j in (l[i]):
		if len(l[j])==0:
			val+=1

	if len(l[i])!=0:
		if val<3:
			flag=0

if flag:
	print 'Yes'
else:
	print 'No'