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:
def dfs(s,prev_leaf,leaf):
    #print(s,prev_leaf,leaf)
    if prev_leaf:
        leaf_child[s]+=1
    for x in g[s]:
        dfs(x,leaf,False)
n=int(input())
leaf_child=[0]*(n+1)
g={x:set() for x in range(1,n+1)}
is_leaf=[1]*(n+1)
for x in range(n-1):
    s=int(input())
    g[x+2].add(s)
    is_leaf[s]=0
for x in range(2,n+1):
    if g[x] != set() and is_leaf[x]:
        dfs(x,False,True)
b=True
for x in range(1,n+1):
    if not is_leaf[x]:
        if leaf_child[x] < 3:
            b=False
print("Yes" if b else "No")