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:
def bfs(s, v, l, r) :
	q = []
	q.append(s)
	while q :
		p = q.pop()
		c = 0
		for i in l[p] :
			if v[i] == 0 :
				if len(l[i]) == 0 :
					c += 1
				else :
					q.append(i)
				v[i] = 1
		#print c
		if c < 3 :
			return 0
	return 1


	
n = input()
l, v = [], []
l.append([])
v.append(0)
for i in range (1, n+1) :
	l.append([])
	v.append(0)
for i in range (1,n) :
	k = input()
	l[k].append(i+1)
p = 0
r = 0

if bfs(1, v, l, r) == 1 :
	print "Yes"
else :
	print "No"