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
sys.setrecursionlimit(1000000)
x = int(input())
m = [[] for i in range(x)]
for i in range(x - 1):
	l = int(input())
	m[l - 1] += [i + 2]
for i in range(x):
	if len(m[i]) != 0:
		m[i] += [len(m[i])]
	else:
		m[i] += [-10000]
k = 1
f = 0
def dfs(k,f):
	if m[k - 1][-1] > 2:
		if f != 0:
			m[f - 1][-1] -= 1
		for i in range(m[k - 1][-1]):
			dfs(m[k-1][i], (k))
	elif m[k - 1][-1] == 1 or m[k - 1][-1] == 2:
		m[f][-1] -= 1
	return 0
dfs(k,f)
t = 0
for i in range(x):
	if m[i][-1] < 3 and m[i][-1] > -10000:
		print('No')
		t = 1
		break
if t == 0:
	print('Yes')