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:
class tree:
    def __init__(self, parent, son):
        self.parent = parent
        self.son = son

if __name__ == "__main__":
    n = int(input())
    trees = []
    for i in range(n+1):
        trees.append(tree(0, 0))
    
    p = [0 for i in range(n+1)]
    for k in range(2, n+1):
        node = int(input())
        if not p[node]:
            p[node] = 1
            trees[trees[node].parent].son -= 1
        trees[k].parent = node
        trees[node].son += 1

    flag = 0
    for i in range(1, n):
        if((trees[i].son and trees[i].son < 3) or (p[i] and trees[i].son < 3)):
            flag = 1
            break
    if flag:
        print("No")
    else:
        print("Yes")