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:
if __name__ == '__main__':
    n = int(input())
    nonleaf = [0 for i in range(1010)]
    child = [[] for i in range(1010)]
    leaf = [0 for i in range(1010)]

    def dfs(s):
        cnt = 0
        for chd in child[s]:
            cnt += dfs(chd)
        leaf[s] = cnt
        return 1 - nonleaf[s]

    for i in range(2, n + 1):
        node = int(input())
        child[node].append(i)
        nonleaf[node] = 1

    dfs(1)

    # print(nonleaf[1:n + 1])
    # print(child[1:n + 1])
    # print(leaf[1:n + 1])

    for i in range(1, n + 1):
        if nonleaf[i] and leaf[i] < 3:
            print("No")
            exit()

    print("Yes")