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:

def create_graph(n):
    graph = [[] for x in range(0,n)]
    return graph

def children(node):
    if len(graph[node]) == 0:
        leafs[node] = 1
        return
    for x in graph[node]:
            children(x)

def dfs(node):
    b = True
    count = 0
    for x in graph[node]:
        if leafs[x]:
            count = count + 1
        else:
            b  = b and dfs(x)
    
    if count < 3 :
        b = False
    return b 

nodes = input()
graph = create_graph(int(nodes))

leafs = [0]*(int(nodes))

for x in range(2,int(nodes)+1):
    a = input()
    graph[int(a)-1].append(x-1)

children(0)  

result = dfs(0)

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