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:
# cook your dish here
from sys import stdin,stdout
from collections import Counter
from itertools import permutations
import bisect
import math
I=lambda: map(int,stdin.readline().split())
I1=lambda: stdin.readline()
#(a/b)%m =((a%m)*pow(b,m-2)%m)%m

n=int(I1())
l=[]
for _ in range(n-1):
    x=int(I1())
    l.append(x)
    
s=list(set(l))
#print(l)
for x in s:
    c=0
    for i in range(n-1):
        if x==l[i]:
            if i+2 not in s:
                c+=1 
    if c<3:
        print("No")
        exit()
print("Yes")