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 JAVA):

Solution:
var n=Number(readline())
var N=new Array(n+1)
N[1]={c:[]}
for (var i=0;i<n-1;++i) {
 var p=Number(readline())
 N[i+2]={c:[]}
 N[p].c.push(N[i+2])
}
var check=i=>{
 var l=N[i].c.length;
 if (l) {
  var cc=0; 
  for (var j=0;j<l;++j)
   cc += (!N[i].c[j].c.length)
  return cc>=3;
 } else return true;
}

var a=true;
for (var i=1;i<=n;++i)
 a = a && check(i)
print(a?'Yes':'No')