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;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void sieveOfEratosthenes(int n, bool prime[]) {
  int i, p;
  for (i = 2; i <= n / 2; i++) {
    if (prime[i]) {
      for (p = i * i; p <= n; p += i) prime[p] = false;
    }
  }
}
unsigned long long int power(unsigned long long int n,
                             unsigned long long int p) {
  unsigned long long int m = 1;
  while (p > 0) {
    m *= n;
    p--;
  }
  return m;
}
bool checkSpruce(int n, std::vector<int> adj[]) {
  int count = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < adj[i].size(); ++j) {
      if (adj[adj[i][j]].size() == 0) count++;
    }
    if (count < 3 && adj[i].size() > 0) return false;
    count = 0;
  }
  return true;
}
int main() {
  int t, n, i, k, m, len, u, v;
  cin >> n;
  vector<int> adj[n];
  for (i = 0; i < n - 1; ++i) {
    cin >> u;
    adj[u - 1].push_back(i + 1);
  }
  if (checkSpruce(n, adj))
    cout << "Yes";
  else
    cout << "No";
  cout << endl;
  return 0;
}