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:
def isSpruceNode(nodes, nodeIndex):
	if len(nodes[nodeIndex]) >= 3:
		nLeafChild = 0
		for childIndex in nodes[nodeIndex]:
			if len(nodes[childIndex]) == 0:
				nLeafChild += 1
				if nLeafChild == 3:
					return True
	return False


nNodes = int(input())
nodes = [[] for i in range(nNodes)]
# [root, child, child, child]
for i in range(1, nNodes):
	nodes[int(input()) - 1].append(i)
# print(nodes)

allSpruceNodes = True
for nodeIndex in range(0, nNodes):
	if len(nodes[nodeIndex]) != 0:
		# print(str(nodes[nodeIndex]) + ' - isSpruceNode: ' + str(isSpruceNode(nodes, nodeIndex)))
		if not isSpruceNode(nodes, nodeIndex):
			print('No')
			allSpruceNodes = False
			break
if allSpruceNodes:
	print('Yes')