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:
n=int(input())
dic={}
dic[1]=1
child =[0 for i in range(n+1)]
node=[0 for i in range(n+1)]
for i in range(2,n+1):
  u=(int(input()))
  node[i]=u
  if u not in dic:
    dic[u]=1
for i in range(2,n+1):
  child[node[i]]+=1
  if node[node[i]]==-1:continue
  child[node[node[i]]]-=1
  node[node[i]]=-1
f="Yes"
'''print node
print child
print dic'''
for i in  range(1,n+1):
  if child[i]<3 and i in dic:
    f="No"
    break
#if child[1]<3:f="No"
print f