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:

import sys

def read():
    N = int(sys.stdin.readline())
    adjlist = [[] for i in range(N)]
    for i in range(1, N):
        u = int(sys.stdin.readline())-1
        v = i
        adjlist[u].append(v)
    return adjlist

def is_spruce(adjlist, u):
    leaf_cnt = 0
    for v in adjlist[u]:
        if len(adjlist[v]) > 0:
            if not is_spruce(adjlist, v):
                return False
        else:
            leaf_cnt += 1
    if leaf_cnt == 0:
        return len(adjlist[u]) == 0
    else:
        return leaf_cnt >= 3



if __name__ == '__main__':
    adjlist = read()
    ret = is_spruce(adjlist, 0)
    if ret:
        print('Yes')
    else:
        print('No')