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:
from sys import stdin,stdout
#stdin.readline()
#int(stdin.readline())
#stdout.write()
#raw_input()
#map(int,raw_input().split())
#map(int,stdin.readline().split())
n=int(stdin.readline())
d={}
for _ in xrange(n-1):
    t=int(stdin.readline())
    try:
        d[t].append(_+2)
    except KeyError:
        d[t]=[_+2]
z=[1]
while len(z)!=0:
    t=z.pop()
    q=d[t]
    l=len(q)
    if l<3:
        print "No"
        exit()
    r=0
    for i in q:
        try:
            x=d[i]
            z.append(i)
        except KeyError:
            r+=1
    if r<3:
        print "No"
        exit()
print "Yes"