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 PYTHON3):

Solution:
'''
author sefat e mahadi
'''
class Node():
	def __init__(self,index):
		self.parent =None
		self.ancestor =[]
		self.end =1
		self.index =index
narr =[]
n =int(input())
for i in range(n):
	narr.append(Node(i))
for i in range(1,n):
	par =int(input())-1
	narr[i].parent =narr[par]
	narr[par].ancestor.append(narr[i])
	narr[par].end =0
'''
for i in range(n):
	print (narr[i].end,narr[i].index)
'''
varr =[0]*n
def f(node):
	if (node == None):
		return
	if (varr[node.index]):
		return
	counter =0
	for a in node.ancestor:
		if (a.end):
			counter+=1
	if (counter < 3):
		print ('No')
		quit()
	varr[node.index] =1
	f(node.parent)
for node in narr:
	if (node.end):
		f(node.parent)
print ('Yes')