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())
l=[0]
for i in range(n-1):
    x=int(input())
    l.append(x)
#print(*l)
#l.sort()
a=[]
for i in range(1,n):
    if l[i] not in a:
        a.append(l[i])
#print(*a)
for i in range(len(a)):
    l[a[i]-1]=0
#print(*l)    
l.sort()
#print(*l)
freq = {} 
v=[]
for items in l: 
    freq[items] = l.count(items) 
for key, value in freq.items(): 
    v.append([key,value])
#print(len(v)-1)    
x=0
for i in range(len(v)):
    if v[i][0]>0:    
        if v[i][1]<3 or len(a)!=len(v)-1:
            print("NO")
            x=1
            break
if x==0:
    print("YES")