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:
# Tree size
n = int(input())

# Define tree and root
tree = dict()
tree[0] = [0, True, 0]
is_pruce = 'Yes'

# Build tree
for i in range(1, n):
    tree[i] = [int(input()), True, 0]

    parent = tree.get(i)[0] - 1

    # Register non-leaf status
    if tree[parent][1] is True:
        tree[parent][1] = False

# Parse tree to compute leaf-child
for i in range(1, n):
    parent = tree.get(i)[0] - 1

    # Compute children
    if tree[i][1] is True:
        tree[parent][2] += 1

for values in tree.values():
    if values[1] is False and values[2] <= 2:
        is_pruce = 'No'
        break

print(is_pruce)