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:
from collections import deque
n = int(input())
graph = []
for i in range(n+1):
    graph.append([])
parent = [-1, -1]

for j in range(2, n+1):
    p = int(input())
    parent.append(p)
    graph[p].append(j)

def bfs(graph, node, visited, leafs, parent):
    q = deque()
    visited[node] = True
    q.append(node)
    while q:
        node = q.popleft()
        if not graph[node]:
            leafs[parent[node]] += 1
        else:
            for i in graph[node]:
                if not visited[i]:
                    visited[i] = True
                    q.append(i)

leafs = [0]*(n+1)
bfs(graph, 1, [False]*(n+1), leafs, parent)

for i in range(len(leafs)):
    if graph[i]:
        if leafs[i] < 3:
            print('NO')
            exit()
print('YES')