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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<int> > adj(n, vector<int>(n, 0));
  set<int> s;
  int x;
  for (int i = 2; i < n + 1; i++) {
    cin >> x;
    adj[x - 1][i - 1] = 1;
    s.insert(x - 1);
  }
  stack<int> st;
  int i = 0;
  int j;
  int count = 0;
  while (i < n) {
    if (s.find(i) == s.end()) {
      i++;
      continue;
    }
    for (j = 0; j < n; j++) {
      if (adj[i][j] == 1 && (s.find(j) == s.end())) {
        count++;
      }
    }
    if (count < 3) {
      cout << "No" << endl;
      return 0;
    }
    count = 0;
    i++;
  }
  cout << "Yes" << endl;
  return 0;
}