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:
from sys import stdin
lines = stdin.readlines()
n = int(lines[0])
dikte = {}

for i in range(1,n):
    c = int(lines[i])
    if c not in dikte:
        dikte[c] = [i+1]
    else:
        dikte[c].append(i+1)

def isleaf(x,dikte):
    nonleaf = []
    leaf = []
    for i in range(1,x+1):
        if i in dikte:
            nonleaf.append(i)
        else:
            leaf.append(i)
    return (nonleaf,leaf)
x = isleaf(n,dikte)[0]
y = isleaf(n,dikte)[1]
def final(x,y,dikte):
    
    for i in x:
        summ = 0   
        for j in dikte[i]:
            if j in y and j not in x:
                summ += 1
        if summ < 3:
            return "No"
    return "Yes"
print final(x,y,dikte)