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())

adjList = []
for i in range(0, n):
    adjList.append([])

par = []
for i in range(0, n):
    par.append(-1)

for i in range(0, n-1):
    parent = int(input())
    parent -= 1
    #vertex i + 1 is a child of vertex parent
    adjList[i+1].append(parent)
    adjList[parent].append(i+1)
    par[i+1] = parent

isLeaf = []
for i in range(0, n):
    isLeaf.append(False)

leafCount = []
for i in range(0, n):
    leafCount.append(0)

for i in range(0, n):
    if len(adjList[i]) == 1 and i != 0:
        isLeaf[i] = True
        leafCount[par[i]] += 1

isValid = True

for i in range(0, n):
    #print(isLeaf[i], leafCount[i])
    if isLeaf[i] == False and leafCount[i] < 3:
        isValid = False

if isValid:
    print("Yes")
else:
    print("No")