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:
def has3leaf(v, h):
    ee = map(str, h[v].strip(',').split(','))
    c=len(ee)
    for e in ee:
        if h.get(int(e)):
            c-=1
    if c<3: return 0
    else: return 1


r=int(raw_input().strip())

l=list()
l.append(r)
for n in xrange(r-1):
    l.append(int(raw_input().strip()))

h=dict()
w=1
for i in l[1:]:
    h[i-1] = str(h.get(i-1,'')) + ','+str(w)
    w+=1

r=0
for e in h:
    ln=len(h[e].split(','))
    if ln < 3:
        continue
    r +=has3leaf(e, h)

if r==len(h): print 'Yes'
else: print 'No'