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:
import sys
from collections import *
from string import *
from itertools import *

class node:
    def __init__(self):
        self.child = []
        self.parent = None

n = input()
nodes = [node() for i in range(n)]
for i in range(1,n):
    pi = input()
    nodes[i].parent = nodes[pi-1]
    nodes[pi-1].child.append(nodes[i])

ans = "Yes"
for i in range(n):
    temp = 0
    if len(nodes[i].child)>0:
        #print i,nodes[i].child
        for ch in nodes[i].child:
            #print ch.child
            if len(ch.child)==0:
                temp += 1
            if temp==3:
                break
        #print temp
        if temp<3:
            ans = "No"
            break
print ans