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:

Using python can you solve the prior task?
 Can you provide a Python code to print "Yes" if the tree is a spruce and "No" otherwise?
assistant
#Here is Python code to print "Yes" if the tree is a spruce and "No" otherwise.

class G:
    def __init__(self,v):
        self.v = v
        self.graph = [[] for _ in range(v)]
    def connect(self,u,v):
        self.graph[u].append(v) # u is the parent of v

def isSpruceUtil(g,src):
    leaf = 0
    spruce = True
    for i in range(len(g.graph[src])):
        if len(g.graph[g.graph[src][i]])==0:
            leaf += 1
        else:
            spruce = spruce and isSpruceUtil(g,g.graph[src][i])
    if leaf >= 3 and spruce:
        return True
    else:
        False

def isSpruce(g):
    return isSpruceUtil(g,0)

def solve():
    n = int(raw_input())
    g = G(n)
    for i in range(1,n,1):
        parent = int(raw_input())
        g.connect(parent-1,i)
    if isSpruce(g):
        print("Yes")
    else:
        print("No")
if __name__ == "__main__":
    solve()