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:

class TreeNode:
    def __init__(self, val):
        self.val = val
        self.children = []


class Solution:
    def __init__(self):
        self.root = TreeNode(1)

    def insertIntoTree(self, indexOfParent, val):
        self.insertHelper(self.root, indexOfParent, val)

    def insertHelper(self, curr, indexOfParent, val):
        if not curr:
            return None

        if curr.val == indexOfParent:
            curr.children.append(TreeNode(val))
            return

        for child in curr.children:
            self.insertHelper(child, indexOfParent, val)

    def isSpruce(self, root):
        self.ans = True

        self.dfs(root)

        return self.ans

    def dfs(self, curr):
        if not curr:
            return None

        if not curr.children:
            return True

        leafCount = 0

        for child in curr.children:
            isLeaf = self.dfs(child)
            leafCount += 1 if isLeaf else 0

        if leafCount < 3:
            self.ans = False

        return False


sol = Solution()

n = int(input().strip())

for i in range(n-1):
    indexOfParent = int(input().strip())
    sol.insertIntoTree(indexOfParent, 2+i)

print("Yes" if sol.isSpruce(sol.root) else "No")