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 pprint import pprint as pp
def GI(): return int(raw_input())
def GIS(): return map(int, raw_input().split())

def main():
  n = GI()
  ps = [0, 0] + [GI() for _ in xrange(n-1)]
  is_leaf = [1] * (n + 1)
  leaf_cnt = [0] * (n + 1)

  for p in ps:
    is_leaf[p] = 0

  for i, p in enumerate(ps):
    if is_leaf[i]:
      leaf_cnt[p] += 1

  #print(is_leaf)
  #print(leaf_cnt)

  for i, c in enumerate(leaf_cnt):
    if not i:
      continue
    if not is_leaf[i] and c < 3:
      print('No')
      return
  print('Yes')

main()