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())
i = 2

edge = dict()
leafs = set()

i = 1
while i <= n:
  edge[i] = set()
  i += 1

i = 2
while i <= n:
  parent = int(input()) 
  edge[parent].add(i)
  i += 1

# print (edge)

dlt = list()
for x in edge.keys():
    if len(edge[x]) == 0:
      leafs.add(x)
      dlt.append(x)

for x in dlt:
  del edge[x]

# print(edge)

for x in edge.values():
  cnt = int(0)
  for y in x:
      if y in leafs:
        cnt += 1
  if cnt < 3:
    print("No")
    quit()

print("Yes")
quit()