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())
parents = []
leaves = []
def argsort(seq):
    return [x for x,y in sorted(enumerate(seq), key = lambda x: x[1])]
def sort(seq):
    return [y for x,y in sorted(enumerate(seq), key = lambda x: x[1])]
for i in range(n-1):
    p = int(input())
    parents.append(p)
for l in range(1,n+1):
    if l not in parents:
        leaves.append(l)
indexes = argsort(parents)
parents = sort(parents)
flag = True
i = 0
res = 'Yes'
while flag:
    p = parents[i]
    lchilds = 0
    while parents[i] == p:
        if indexes[i]+2 in leaves:
            lchilds += 1
        if i < n-2:
            i += 1
        else:
            p = -1
    if lchilds < 3:
        res = 'No'
        flag = False
    if i == n-2:
        flag = False
print(res)