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:
n=int(input())
koldet=list()
rod=list()
pravda=1
kollistov=list()
for i in range(1,n+2):
   koldet.append(int(0))
   rod.append(0)
   kollistov.append(0)
for i in range(2,n+1):
   m=int(input())
   koldet[m]+=1
   rod[i]=m
for i in range(1,n+1):
    if koldet[i]==0:
        kollistov[rod[i]]+=1 
for i in range(1,n+1):
    if  kollistov[i]<3 and koldet[i]!=0:
        pravda*=0
if pravda==1:
    print("Yes")
else:
    print("No")