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:
# -*- coding: utf-8 -*-
import sys


def readline(sep=None): return map(int, raw_input().split(sep))
def writeline(seq): print(" ".join(map(str, seq)))


def main():
    n, = readline()
    parent = [int(token) - 1 for token in sys.stdin.readlines()]

    children = [list() for __ in xrange(n)]

    for c, p in enumerate(parent):
        children[p].append(c + 1)

    for ch in children:
        if ch:
            spikes = sum(not bool(children[x]) for x in ch)
            if spikes < 3:
                print "No"
                break
    else:
        print "Yes"


main()