human
Can you provide a Python code to print "Yes" if the tree is a spruce and "No" otherwise given the following:
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:

class
assistant
#Here is Python code to print "Yes" if the tree is a spruce and "No" otherwise.
 CodeforcesTask913BSolution:
    def __init__(self):
        self.result = ''
        self.n = 0
        self.parents = []

    def read_input(self):
        self.n = int(input())
        for x in range(self.n - 1):
            self.parents.append(int(input()))

    def process_task(self):
        child = [[] for x in range(self.n)]
        for x in range(self.n):
            child[self.parents[x - 1] - 1].append(x + 1)
        leaf_child = [[] for x in range(self.n)]
        for x in range(self.n):
            #print(x, len(child[x]))
            if not len(child[x]):
                leaf_child[self.parents[x - 1] - 1].append(x + 1)
                leaf_child[x] = None
        valid = True
        #print(child)
        #print(leaf_child)
        for x in range(self.n):
            if leaf_child[x]:
                if len(leaf_child[x]) < 3:
                    valid = False
                    break
            elif leaf_child[x] == []:
                valid = False
                break
        if not valid:
            self.result = "No"
        else:
            self.result = "Yes"

    def get_result(self):
        return self.result

if __name__ == "__main__":
    Solution = CodeforcesTask913BSolution()
    Solution.read_input()
    Solution.process_task()
    print(Solution.get_result())