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:
"""
8
1
1
1
1
3
3
3
Yes

4
1
1
1
Yes

7
1
1
1
1
2
2
No

8
1
1
1
1
5
5
5
Yes

9
1
1
1
1
2
6
6
6
No

12
1
1
1
2
5
5
5
5
1
2
2
No
"""
from math import sqrt

R = lambda: tuple(map(int, input().split(" ")))


def main():
    n = int(input())
    tree = {}
    leaves, parents = set(), set()
    for i in range(1, n):
        p = int(input())
        tree.setdefault(p, []).append(i + 1)
        parents.add(p)
        leaves.discard(p)
        leaves.add(i + 1)
    for node in parents:
        leaf_count = 0
        for i in tree[node]:
            if i in leaves:
                leaf_count += 1
        if leaf_count < 3:
            print("No")
            return
    print("Yes")


if __name__ == "__main__":
    main()