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())
arr = [[0, -1, True]] + [[0, 0, False]]*(n-1) #score, parent, isParent
for i in range(1, n):
    a = int(input())
    arr[i] = [0, a-1, False]
    if(arr[a-1][2]==True):
        arr[a-1] = [arr[a-1][0]+1, arr[a-1][1], True]
    else:
        if(arr[a-1][1]>=0):
            arr[arr[a-1][1]][0] -= 1
        arr[a-1] = [arr[a-1][0]+1, arr[a-1][1], True]
#print(arr)
for i in range(n):
    if((arr[i][2]==True and arr[i][0]==0) or (arr[i][0]<3 and arr[i][0]>0)):
        #print(i)
        print('No')
        break
else:
    print('Yes')