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())

G= { i:[] for i in range(n) }

for i in range(n-1):
    x=int(input())
    G[x-1].append(i+1)

inv = True

def bfs():
    global inv
    visited=[False for i in range(n)]
    q=[]
    q.append(0)
    while q!=[]:
        x = q.pop(0)
        y = G[x]
        if visited[x]==False:
            if y !=[]:
                count=0
                for i in y:
                    if G[i]==[]:
                        count+=1
                    q.append(i)
                if count<3:
                    inv = False
                    return
            visited[x]=True
    return

bfs()

if inv:
    print("Yes")
else:
    print("No")