human
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:


The second example:


It is not a spruce, because the non-leaf vertex 1 has only 2 leaf children.

The third example:



cla ... Can you provide a Python program to print "Yes" if the tree is a spruce and "No" otherwise?
assistant
#Here is Python program to print "Yes" if the tree is a spruce and "No" otherwise.
ss tree:
    def __init__(self, index):
        self.children = []
        self.isleaf = True
        self.parent = None
        self.index = index

    def add(self, graph):
        self.children.append(graph)
        graph.parent = self
        self.isleaf = False

    def is_spruce(self):
        ctr = 0
        for el in self.children:
            if el.isleaf:
                ctr += 1
            else:
                if not el.is_spruce():
                    return False
        if ctr >= 3:
            return True
        return False

def find_index(root, index):
    if root.index == index:
        return root
    if root.isleaf:
        return False
    for el in root.children:
        a = find_index(el, index)
        if a:
            return a
    return False

root = tree(1)
edges = int(input())
for i in range(2, edges+1):
    find_index(root, int(input())).add(tree(i))

if root.is_spruce():
    print("Yes")
else:
    print("No")