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


def dfs (v):
  if len(next[v]) == 0:
    leaf[v] = 1
  for u in next[v]:
    dfs (u)

n = int(sys.stdin.readline())
next = dict()
for i in range (1, n + 1):
  next[i] = []
for i in range (n - 1):
  m = int(sys.stdin.readline())
  next[m].append (i + 2)

leaf = [0 for i in range (n + 1)]
dfs (1)
ok = 1
for i in next:
  if len (next[i]) == 0:
    continue
  s = 0
  for t in next[i]:
    s += leaf[t]
  if s < 3:
    ok = 0     
print (["No", "Yes"][ok])