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:
from collections import *
import math
from sys import *
inp=stdin.readline
str=str
flag=1
if not flag:
    out=stdout.write
else:
    out=print
    
 
def linp(x):
    return list(map(int,x.split()))
def minp(x):
    return map(int,x.split())
    


n=int(inp())
dic=defaultdict(list)
ii=1
while n-1>0:
    dic[int(inp())].append(ii+1)   
    ii+=1
    n-=1
flag=0
ans={key:0 for key in dic}
for i in dic:
    for j in dic[i]:
        if j not in dic:
            ans[i]+=1

for i,v in ans.items():
    if v<3:
        out("No")
        break
else:
    out("Yes")