#include <bits/stdc++.h>
using namespace std;
const int INF32 = 1E9;
const long long INF64 = 4E18;
const long long M = 1E9 + 7;
const double EPS = 1E-9;
const double PI = 2 * acos(0);
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (T x : p) os << x << " ";
  os << "]" << '\n';
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}" << '\n';
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const map<Tk, Tv>& p) {
  os << "{ ";
  for (pair<Tk, Tv> x : p) os << x << " ";
  os << "}" << '\n';
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const pair<Tk, Tv>& p) {
  os << "{" << p.first << ',' << p.second << "}";
  return os;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("tupni.txt", "r")) {
    freopen("tupni.txt", "r", stdin);
    freopen("tuptuo.txt", "w", stdout);
  }
  int n;
  cin >> n;
  vector<vector<long long> > g(n + 1);
  for (int i = 2; i <= n; ++i) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  bool flag = true;
  for (int i = 1; i <= n; ++i) {
    int cnt = 0;
    for (int j = 0; j < (int)g[i].size(); ++j) {
      int v = g[i][j];
      if ((int)g[v].size() == 0) {
        cnt++;
      }
    }
    if ((int)g[i].size() != 0) {
      if (cnt < 3) {
        flag = false;
        break;
      }
    }
  }
  if (flag)
    cout << "Yes" << '\n';
  else
    cout << "No" << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
const int INF32 = 1E9;
const long long INF64 = 4E18;
const long long M = 1E9 + 7;
const double EPS = 1E-9;
const double PI = 2 * acos(0);
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& p) {
  os << "[ ";
  for (T x : p) os << x << " ";
  os << "]" << '\n';
  return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& p) {
  os << "{ ";
  for (T x : p) os << x << " ";
  os << "}" << '\n';
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const map<Tk, Tv>& p) {
  os << "{ ";
  for (pair<Tk, Tv> x : p) os << x << " ";
  os << "}" << '\n';
  return os;
}
template <typename Tk, typename Tv>
ostream& operator<<(ostream& os, const pair<Tk, Tv>& p) {
  os << "{" << p.first << ',' << p.second << "}";
  return os;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  if (fopen("tupni.txt", "r")) {
    freopen("tupni.txt", "r", stdin);
    freopen("tuptuo.txt", "w", stdout);
  }
  int n;
  cin >> n;
  vector<vector<long long> > g(n + 1);
  for (int i = 2; i <= n; ++i) {
    int p;
    cin >> p;
    g[p].push_back(i);
  }
  bool flag = true;
  for (int i = 1; i <= n; ++i) {
    int cnt = 0;
    for (int j = 0; j < (int)g[i].size(); ++j) {
      int v = g[i][j];
      if ((int)g[v].size() == 0) {
        cnt++;
      }
    }
    if ((int)g[i].size() != 0) {
      if (cnt < 3) {
        flag = false;
        break;
      }
    }
  }
  if (flag)
    cout << "Yes" << '\n';
  else
    cout << "No" << '\n';
  return 0;
}

```