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 PYTHON):

Solution:
def IsLeaf(edges,i):
    if edges[i] == []:
        return True
    else:
        return False

def Has3LeafChilds(edges,i):
    count = 0
    for j in edges[i]:
        if IsLeaf(edges,j):
            count += 1
    
    if count >= 3:
        return True
    else:
        return False


def IsSpruce(edges):
    for i in edges:
        if edges[i] != [] and not Has3LeafChilds(edges,i):
            return False
        
    return True

def Spruce():
    n = int(raw_input())
    edges = {}
    for i in range(1,n+1):
        edges[i] = []
    for i in range(2,n+1):
        index = int(raw_input())
        edges[index].append(i)
    
    if IsSpruce(edges):
        print 'Yes'
    else:
        print 'No'

Spruce()