human
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:


The second example:


It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.

The third example:

def m
Write me a Python code to print "Yes" if the tree is a spruce and "No" otherwise?
assistant
#Here is Python code to print "Yes" if the tree is a spruce and "No" otherwise.
ain():
    n = int(input())
    nodes = list()
    nodes.append(ThisTree())
    for i in range(1, n):
        p = int(input())
        nodes.append(ThisTree())
        nodes[p - 1].childs.add(i)
        nodes[p - 1].is_vertex = True

    for node in nodes:
        if not node.childs:
            continue

        count_non_vertex = 0
        for child in node.childs:
            if not nodes[child].is_vertex:
                count_non_vertex += 1

        if count_non_vertex < 3:
            print('No')
            return

    print('Yes')

class ThisTree():
    def __init__(self):
        self.childs = set()
    is_vertex = False

if __name__ == '__main__':
    main()