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;
bool compare(long long a, long long b) { return a > b; }
bool dfs(long long curr, vector<vector<long long>> &adj) {
  if (adj[curr].empty()) {
    return true;
  }
  long long leaf = 0;
  for (long long neighbour : adj[curr]) {
    if (adj[neighbour].empty()) {
      leaf++;
    }
  }
  if (leaf >= 3) {
    bool ans = true;
    for (long long neighbour : adj[curr]) {
      if (!dfs(neighbour, adj)) {
        ans = false;
      }
    }
    return ans;
  } else {
    return false;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t-- > 0) {
    long long n;
    cin >> n;
    vector<vector<long long>> adj(n + 1);
    for (long long i = 0; i < n - 1; i += 1) {
      long long temp;
      cin >> temp;
      adj[temp].push_back(i + 2);
    }
    if (dfs(1, adj)) {
      cout << "Yes\n";
    } else {
      cout << "No\n";
    }
  }
  return 0;
}