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:
import sys
from collections import deque
s = sys.stdin 

n = int(s.readline().split()[0])

tree = [[] for j in range(n)]

for j in range(n - 1):
    child = j + 1
    parent = int(s.readline().split()[0]) - 1
    tree[parent].append(child)


def bfs(tree):
    queue = deque([0])
    result = True
    while len(queue) > 0:
        u = queue.popleft()
        leaf_count = 0
        for v in tree[u]:
            if len(tree[v]) == 0:
                leaf_count += 1
            else:
                queue.append(v)
        result = result and leaf_count > 2
        if not result:
            return result
    return result


result = bfs(tree)


if result:
    print('Yes')
else:
    print('No')