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:
"""
NTC here
"""
from sys import setcheckinterval, stdin, setrecursionlimit
setcheckinterval(1000)
setrecursionlimit(10**7)
 
# print("Case #{}: {} {}".format(i, n + m, n * m))
 
 
def iin(): return int(stdin.readline())
 
 
def lin(): return list(map(int, stdin.readline().split()))


n=iin()

adj=[[] for i in range(n)]
for i in range(n-1):
    b=iin()
    adj[b-1].append(i+1)

lf=set()
for i in range(n):
    if adj[i]==[]:
        lf.add(i)
for i in range(n):
    ch=0
    if i in lf:
        continue
    for j in adj[i]:
        if j in lf:
            ch+=1
    #print(adj[i],ch,lf)
    if ch<3:
        print('No')
        exit()
print('Yes')