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;
vector<int> adj[1234];
int parent[1234];
int visited[1234];
int main() {
  int n;
  cin >> n;
  int number;
  queue<int> qq;
  for (int i = 2; i <= n; i++) {
    cin >> number;
    parent[number] = 1;
    adj[number].push_back(i);
  }
  qq.push(1);
  while (!qq.empty()) {
    int q = qq.front();
    qq.pop();
    int cnt = 0;
    for (int i = 0; i < adj[q].size(); i++) {
      int a = adj[q][i];
      if (!parent[a])
        cnt++;
      else
        qq.push(a);
    }
    if (cnt < 3) {
      printf("No");
      return 0;
    }
  }
  printf("Yes\n");
}