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:
# -*- coding: utf-8 -*-
"""
Created on Tue Nov 17 13:44:48 2020

@author: pctab
"""
import sys
sys.setrecursionlimit(99999999)
from collections import defaultdict
parent=defaultdict(int)
children=defaultdict(set)
n=int(input())
arr=[]
for i in range(n-1):
    x=int(input())
    children[x].add(i+2)
    parent[i+2]=x
leaf=set()
for i in range(1,n+1):
    if not children[i] and parent[i]!=0:
        leaf.add(i)
f=True
for x in children:
    if x in leaf:
        continue
    c=0
    for y in children[x]:
        if y in leaf:
            c+=1
    if c<3:
        f=False
        break
if f:
    print('Yes')
else:
    print('No')