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:
import sys
input=sys.stdin.readline
def main():
	n=int(input())
	A=[None]+[[] for _ in range(n)]
	for i in range(2,n+1) :
		p=int(input())
		A[p].append(i)
	def isLeaf(u) :
		return A[u]==[]
	def isSpurce(u) :
		if isLeaf(u) : return True 
		cnt = 0
		for v in A[u] : 
			if isLeaf(v) : cnt+=1 ; continue
			if not isSpurce(v) : return False
		return cnt > 2 
	print ( 'YES' if isSpurce(1) else 'No')
main()