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 find(i,p):
	List = True
	for j in range(n):
		if a[i][j]==1:
			List = False
			find(j,i)
	if List:
		was[p]+=1
	else:
		was[i]+=1
n = int(input())
a = []
was = [0]*n
for i in range(n):
	a.append([0]*n)
for i in range(n-1):
	p = int(input())
	a[p-1][i+1]=1
find(0,0)
ans = True
for i in range(n):
	if was[i]==1 or was[i]==2 or was[i]==3:
		ans = False
		break
if ans:
	print("Yes")
else:
	print('No')