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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  long long int n;
  cin >> n;
  vector<pair<bool, vector<long long int> > > nodes(1007);
  for (long long int i = 0; i < n - 1; i++) {
    long long int t;
    cin >> t;
    nodes[t].first = true;
    nodes[t].second.push_back(i + 2);
  }
  long long int cnt = 0;
  bool flag = true;
  for (long long int i = 0; i < n - 1; i++) {
    cnt = 0;
    if (nodes[i].first) {
      for (long long int j = 0; j < nodes[i].second.size(); j++) {
        if (!nodes[nodes[i].second[j]].first) {
          ++cnt;
        }
      }
      if (cnt < 3) {
        cout << "No" << endl;
        return 0;
      }
    }
  }
  cout << "Yes" << endl;
  return 0;
}