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())
dick = {}
boy = 1
tree = [int(input()) for i in range(n-1)]
#tree.sort()
heads = set()
for i in tree:
    boy=boy+1
    dick[boy]=i
    if i not in heads:
        heads.add(i)
    if i in dick:
        dick.pop(i)
#print(dick)
tree={}
for i in heads:
    tree[i]=0
for i in dick:
    if dick[i] in tree:
        tree[dick[i]]+=1
    else:
        tree[dick[i]]=1
#print(tree)
def chk(tree):
    for i in tree:
        if tree[i]<3:
            return "No"
    return "Yes"
                                                                                                        
print(chk(tree))