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 sys import stdin,stdout
from collections import defaultdict
def dfs(src,par):
    flag=0
    for neigh in g[src]:
        if neigh==par:continue
        dp[src]+=dfs(neigh,src)
        flag=1
    if flag==0:return 1
    return 0
for _ in range(1):#int(stdin.readline())):
    n=int(stdin.readline())
    # n,k=list(map(int,stdin.readline().split()))
    g=defaultdict(list)
    for i in range(2,n+1):
        nn=int(stdin.readline())
        g[nn]+=[i]
        g[i]+=[nn]
    # print(g)
    ans='Yes'
    dp=[0]*(1+n)
    dfs(1,-1)
    # print(dp)
    for i in range(1,n+1):
        if dp[i]==0 and len(g[i])>1:
            ans='No'
            break
        if dp[i] in [1,2]:
            ans='No'
            break
    print(ans)

# 13
# 1
# 2
# 2
# 2
# 1
# 6
# 6
# 6
# 1
# 10
# 10
# 10