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:
from sys import stdin, stdout

#  Author: Andrés Cimadamore(@acimadamore)
#  Judge: Code Forces
#  Problem: Christmas Spruce
#  URL: http://codeforces.com/contest/913/problem/B
#  Topics: tree, tree traversal

def is_spruce(root):
    if(len(tree[root]) < 3):
        return False

    leaves = 0

    for t in tree[root]:
        if t in tree and tree[t]:
            if not is_spruce(t):
                return False
        else:
            leaves+=1

    return leaves >= 3

tree = {} 

n = int(stdin.readline())

for i in range(2, n+1):
    p = int(stdin.readline())

    if p in tree:
        tree[p].append(i)
    else:
        tree[p] = [i]

stdout.write(str("Yes" if is_spruce(1) else "No") + "\n")