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:
# B

n = int(input())    # number of vertices
p = [0, 0]
for i in range(n-1):
    p.append(int(input()))  # p[i] is vertex i's parent

num_child = [0 for i in range(n+1)]   # number of children of vertex i

for i in range(1, n+1):
    num_child[i] = p.count(i)

flag = 0
for i in range(1, n+1):
    if 0 < num_child[i] < 3:
        print("No")
        flag = 1
        break
    elif num_child[i] >= 3:
        num_leaf = 0
        for j in range(1, n+1):
            if p[j] == i and num_child[j] == 0:
                num_leaf += 1
        if num_leaf < 3:
            print("No")
            flag = 1
            break

if flag == 0:
    print("Yes")