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:
def is_spruce(adjacency_list):
    for v in adjacency_list.keys():
        # to check if the vertex is a leaf node or not
        if (len(adjacency_list[v]) != 0):
            count_of_leaf_children = 0
            for c in adjacency_list[v]:
                if(len(adjacency_list[c]) == 0):
                    count_of_leaf_children += 1
            if (count_of_leaf_children < 3):
                print("No")
                return None
    print("Yes")
    return None


no_of_nodes = int(input())
adj_list = {vertices: [] for vertices in range(1, no_of_nodes+1)}
for v in range(2, no_of_nodes+1):
    parent_vertex = int(input())
    adj_list[parent_vertex].append(v)
is_spruce(adj_list)