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;
queue<int> q;
void bfs(vector<int> adj[], int n) {
  if (q.empty()) {
    cout << "YES\n";
    exit(0);
  }
  int current = q.front();
  q.pop();
  if (adj[current].size() != 0) {
    int numleaf = 0;
    for (int j = 0; j < adj[current].size(); j++) {
      if (adj[(adj[current][j])].size() == 0)
        numleaf++;
      else
        q.push(adj[current][j]);
    }
    if (numleaf < 3) {
      cout << "NO\n";
      exit(0);
    }
  }
  bfs(adj, n);
}
int main() {
  int n;
  cin >> n;
  int a[n + 1];
  a[0] = -999;
  a[1] = -999;
  for (int i = 2; i <= n; i++) cin >> a[i];
  vector<int> adj[n + 1];
  for (int i = 2; i <= n; i++) adj[a[i]].push_back(i);
  q.push(1);
  bfs(adj, n);
}