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())

tree = {}

for i in range( 2,n+1 ):
  v = int( input() )
  tree[v] = tree.get(v, [])
  tree[v].append(i)



for i in range(1,n+1):
  # print("tree %d" % i)
  # print( tree.get(i, []) )

  children = tree.get(i, [])

  if (len(children) == 0):
    continue

  if len( children ) < 3:
    print("No")
    exit()

  nofne = 0
  for nei in children:
    if len(tree.get(nei, [])) == 0:
      nofne = nofne + 1

  # print("no of nei: %d" % nofne)
  if nofne < 3:
    print("No")
    exit()


print("Yes")