#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long inf = 1e18;
long long m_inf = INT_MIN;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  std::cout.unsetf(std::ios::fixed);
  std::cout.precision(9);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<vector<long long> > adj(n);
    for (long long i = 1; i < n; i++) {
      long long x;
      cin >> x;
      x--;
      adj[x].push_back(i);
    }
    bool flag = true;
    for (long long i = 0; i < n; i++) {
      long long count = 0;
      if (!adj[i].size()) continue;
      for (long long j = 0; j < adj[i].size(); j++) {
        if (adj[adj[i][j]].size() == 0) count++;
      }
      if (count < 3) {
        flag = false;
        break;
      }
    }
    if (flag)
      cout << "Yes\n";
    else
      cout << "No\n";
  }
}

### Prompt

Please create a solution in cpp to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long inf = 1e18;
long long m_inf = INT_MIN;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  std::cout.unsetf(std::ios::fixed);
  std::cout.precision(9);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<vector<long long> > adj(n);
    for (long long i = 1; i < n; i++) {
      long long x;
      cin >> x;
      x--;
      adj[x].push_back(i);
    }
    bool flag = true;
    for (long long i = 0; i < n; i++) {
      long long count = 0;
      if (!adj[i].size()) continue;
      for (long long j = 0; j < adj[i].size(); j++) {
        if (adj[adj[i][j]].size() == 0) count++;
      }
      if (count < 3) {
        flag = false;
        break;
      }
    }
    if (flag)
      cout << "Yes\n";
    else
      cout << "No\n";
  }
}

```