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:
'''
	Arpan Tarkas
	IIITDM Jabalpur
'''
 
import sys,math
from collections import defaultdict as df
from bisect import bisect_left as bl

raw_input = lambda: sys.stdin.readline().rstrip()
r_s = lambda: raw_input().strip()                   #read str
r_ss = lambda: raw_input().split()          #read stringss
r_i = lambda: int(raw_input())              #read int
r_is = lambda: map(int, raw_input().split())#read ints
r_f = lambda: float(raw_input())            #read float
r_fs = lambda: map(Decimal, raw_input().split()) #read floats
 
#for efficiently taking lot of space separated queries on each line, use this.
r_qs = lambda: map(lambda s:map(int, s.split(' ')), sys.stdin.readlines())


if __name__ == '__main__':
    #start here
    n=r_i()
    dd=df(list)
    for i in xrange(n-1):
        dd[r_i()].append(i+2)
    ddk=dd.keys()
    ddi=dd.items()
    f=0
    for i in xrange(len(ddi)):
        c=0
        for j in xrange(len(ddi[i][1])):
            if ddi[i][1][j] not in ddk:
                c+=1
        if c<3:
            print 'No'
            f=1
            break
    if f==0:
        print 'Yes'