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:
from sys import stdin, stdout

label = 1


def dfs(v):
    global label
    
    if not len(vertices[v]):
        return 1
    else:
        cnt = 0
        
        for u in vertices[v]:
            cnt += dfs(u)
        
        if cnt < 3:
            label = 0
        
        return 0


n = int(stdin.readline())
vertices = [[] for i in range(n + 10)]
parents = [0 for i in range(n + 10)]

for i in range(1, n):
    parents[i + 1] = int(stdin.readline())
    vertices[parents[i + 1]].append(i + 1)


dfs(1)
if label:
    stdout.write('Yes')
else:
    stdout.write('No')