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 PYTHON):

Solution:
def dfs(c, f, t):
    ans, cnt = 0, 0
    if len(t[c]) == 1: return ans, 1
    for i in t[c]:
        if i == f: continue
        x, y = dfs(i, c, t)
        if x: ans = 1
        cnt += y
    if cnt < 3: return 1, cnt
    return ans, 0
n = input()
t = [[] for i in xrange(1005)]
for i in xrange(2, n+1):
    x = input()
    t[i].append(x)
    t[x].append(i)
ans, cnt = dfs(1, -1, t)
print 'No' if ans else 'Yes'