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:
n=int(input())
a=[0]*(n+1)
for  i in range (2,n+1):
    a[i]=int(input())
b=set(a[2:])


b=list(b)
#print(b)
#print(a)

ans=[]
for j in range(1,n+1):
    if j not in b:
#        print(j)
        ans.append(a[j])
#print(ans)
c=[0]*(n+1)
for i in range(len(ans)):
    c[ans[i]]+=1
#print(c)
flag=0
ll=0
for i in range(n+1):
    if c[i]!=0:
        ll+=1
    if c[i]!=0 and c[i]<=2:
        flag+=1
if flag!=0 or ll!=len(b):
    print("NO")
else:
    print("YES")