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 PYTHON):

Solution:
import sys
v = int(raw_input())

node_stat = { 1 : True }
node_map = {}
node_c = 2
for x in range(v-1):
    p = int(raw_input())
    if p not in node_map:
        node_map[p] = []
    node_map[p].append(node_c)
    node_stat[p] = True
    node_c = node_c + 1

p_c = 0

for i,j in node_map.iteritems():
    if i in node_stat:
        p_c = p_c + 1
        cnt = 0
        for cs in j:
            if cs not in node_stat:
                cnt = cnt + 1
        if cnt < 3:
            print "No"
            sys.exit()

if p_c == 0:
    print "No"
else:
    print "Yes"