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())
a = [1]*(n+1)
a[0] = 0
b = [0]*(n+1)
b[0] = -1
b[1] = 1
p = {}
fl = 1
for i in range(2, n+1):
    a[i] = int(input())
    b[a[i]] = 1
    p[a[i]] = []
for i in range(2, n+1):
    p[a[i]].append(i)
c = 0
for keys in p:
    z = p[keys]
    c = 0
    for i in z:
        if b[i] == 0:
            c = c+1
    if c < 3:
        fl = 0
        break
if fl == 0:
    print("No")
else:
    print("Yes")