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:
n = int(input())
adj=[[] for i in range(n+5)]
v=[0]*(n+5)
global count
count=[0]*(n+5)
l=[0]*(n+5)

def dfs(adj,u,v,count):
    v[u]=1
    l[u]=-1
    for i in range(len(adj[u])):
        if len(adj[adj[u][i]])==1:
            # is a leaf node
            l[adj[u][i]]=1
            count[u]+=1
        elif v[adj[u][i]]==0:
            l[adj[u][i]]=1
            dfs(adj,adj[u][i],v,count)

def check(count):
    for i in range(len(count)):
        if count[i]>0 and count[i]<3:
            return False
        if count[i]==0 and l[i]==-1:
            return False
    return True

for i in range(2,n+1):
    u=int(input())
    adj[i].append(u)
    adj[u].append(i)
dfs(adj,1,v,count)
if check(count):
    print("Yes")
else:
    print("No")