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 Sat Sep  7 20:11:24 2019

@author: avina
"""

def dfs(u, visted,d,h):
    
    visted[u-1] = True
    t = 0
    for i in d[u]:
        if visted[i - 1] == False:
            if len(d[i]) == 0:
                t+=1
            dfs(i, visted,d,h)
    if t<3 and len(d[u]) >0:
        h[u] = False
            
from collections import defaultdict
d = defaultdict(list)
n = int(input())

for i in range(n-1):
    a = int(input())
    d[a].append(i+2)
h = [True]*n
visted = [False]*n
dfs(1,visted,d,h)
if sum(h) <n:
    print('NO')
else:
    print('YES')