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:
import math
from collections import defaultdict  as dt
from sys import stdin
inp    = lambda : stdin.readline().strip()#input()
intinp = lambda : int(inp())#int(input())
mod = int(1e9)+7
inf = float("inf")
listinp= lambda : list(map(int,stdin.readline().split()))#list(input())
mapinp = lambda : map(int,stdin.readline().split())#map(input())


#####  Code Goes here   #####################
for _ in range(1):
    n=intinp()
    b=[[] for i in range(n+1)]
    for i in range(2,n+1):
        b[intinp()].append(i);
    c=True
    for i in range(n+1):
        d=0
        if(len(b[i])==0): continue
        for j in b[i]:
            if(len(b[j])==0): d+=1
        if(d<3):    
            c=False
            break;
    if(c):print("YES")
    else:print("NO")