#include <bits/stdc++.h>
using namespace std;
template <class T>
void print(T t) {
  cerr << t;
}
template <class T1, class T2>
void print(pair<T1, T2> p) {
  cerr << "{ ";
  print(p.first);
  cerr << ", ";
  print(p.second);
  cerr << " }";
};
template <class T>
void print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
};
template <class T>
void print(set<T> s) {
  cerr << "[ ";
  for (T i : s) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
};
template <class T>
void print(multiset<T> ms) {
  cerr << "[ ";
  for (T i : ms) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
};
template <class T1, class T2>
void print(map<T1, T2> m) {
  cerr << "[ ";
  for (auto i : m) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void print1d(T a[], long long n) {
  cerr << "a ="
       << " ";
  cerr << "{ ";
  for (long long i = 0; i < n; i++) {
    cerr << a[i];
    cerr << " ";
  }
  cerr << "}\n";
}
void FAST_IO() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void IOE() {}
void Go() {
  long long n = 0;
  cin >> n;
  long long h[1001] = {0};
  vector<long long> graph[1001];
  for (long long i = 2; i <= n; i++) {
    long long u = 0;
    cin >> u;
    graph[u].push_back(i);
    h[u]++;
  }
  bool is = 1;
  for (long long i = 1; i <= n; i++) {
    if (h[i] == 0) continue;
    if (h[i] >= 1 and h[i] < 3) {
      is = 0;
      break;
    }
    for (long long j = 0; j < graph[i].size(); j++) {
      if (h[graph[i][j]] != 0) {
        h[i]--;
      }
      if (h[i] < 3) {
        is = 0;
        break;
      }
    }
  }
  if (is)
    cout << "Yes\n";
  else
    cout << "No\n";
}
int32_t main() {
  FAST_IO();
  long long t = 1;
  while (t--) {
    Go();
  }
  return 0;
}

### 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;
template <class T>
void print(T t) {
  cerr << t;
}
template <class T1, class T2>
void print(pair<T1, T2> p) {
  cerr << "{ ";
  print(p.first);
  cerr << ", ";
  print(p.second);
  cerr << " }";
};
template <class T>
void print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
};
template <class T>
void print(set<T> s) {
  cerr << "[ ";
  for (T i : s) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
};
template <class T>
void print(multiset<T> ms) {
  cerr << "[ ";
  for (T i : ms) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
};
template <class T1, class T2>
void print(map<T1, T2> m) {
  cerr << "[ ";
  for (auto i : m) {
    print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void print1d(T a[], long long n) {
  cerr << "a ="
       << " ";
  cerr << "{ ";
  for (long long i = 0; i < n; i++) {
    cerr << a[i];
    cerr << " ";
  }
  cerr << "}\n";
}
void FAST_IO() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
void IOE() {}
void Go() {
  long long n = 0;
  cin >> n;
  long long h[1001] = {0};
  vector<long long> graph[1001];
  for (long long i = 2; i <= n; i++) {
    long long u = 0;
    cin >> u;
    graph[u].push_back(i);
    h[u]++;
  }
  bool is = 1;
  for (long long i = 1; i <= n; i++) {
    if (h[i] == 0) continue;
    if (h[i] >= 1 and h[i] < 3) {
      is = 0;
      break;
    }
    for (long long j = 0; j < graph[i].size(); j++) {
      if (h[graph[i][j]] != 0) {
        h[i]--;
      }
      if (h[i] < 3) {
        is = 0;
        break;
      }
    }
  }
  if (is)
    cout << "Yes\n";
  else
    cout << "No\n";
}
int32_t main() {
  FAST_IO();
  long long t = 1;
  while (t--) {
    Go();
  }
  return 0;
}

```