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:

'''
tt = int(raw_input())
a = map(int, raw_input().split())
for i in xrange(tt):
a = map(int, (' '.join(n for n in raw_input())).split())
'''

n = int(raw_input())
tree = {}
child = {}
for t in xrange(1, n):
    tree[t+1] = int(raw_input())

for c in tree:
    p = tree[c]
    if p in child:
        child[p].append(c)
    
    else:
        child[p] = [c]

leaf = [0]*(n+1)
for i in xrange(1, n+1):

    if i not in child:
        leaf[i] = 1

count = [0]*(n+1)
for c in tree:
    p = tree[c]
    if leaf[c] == 1:
        count[p] += 1
    

# print child
# print count, leaf
ans = "Yes"
for i in xrange(1, n+1):

    if i in child:
        if count[i] >= 3:
            ans = "Yes"
        
        else:
            ans = "No"
            break
    
print ans