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:
import sys

n = int(raw_input())

parents = []

for i in range(n-1):
	parents.append(int(raw_input()))

non_leaf = {}

leaf = []

non_leaf[1] = 0

for i in range(0,len(parents)):
	non_leaf[parents[i]] = 0

# print non_leaf

for i in range(len(parents)):
	if i+2 not in non_leaf:
		leaf.append(i+1)

# print leaf

for i in leaf:
	non_leaf[parents[i-1]] += 1

# print non_leaf

flag = 0

for i in non_leaf:
	if non_leaf[i] < 3:
		flag = -1
		break

if flag == 0:
	print 'Yes'
else:
	print 'No'