#include <bits/stdc++.h>
using namespace std;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
int main(void) {
  vector<vector<int> > tree(1000);
  vector<bool> nonleaf(1000);
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    tree[i].push_back(0);
  }
  for (int i = (0); i < (n - 1); ++i) {
    int parent;
    cin >> parent;
    tree[parent - 1].push_back(i + 1);
  }
  for (int i = (0); i < (n - 1); ++i) {
    if (tree[i].size() > 1) {
      cerr << "i"
           << " = " << (i) << endl;
      ;
      nonleaf[i] = true;
    }
  }
  bool flag = true;
  for (int i = (0); i < (n); ++i) {
    if (nonleaf[i]) {
      cerr << "i"
           << " = " << (i) << " (L" << 75 << ")"
           << " "
           << "<stdin>" << endl;
      ;
      cerr << "tree[i].size()"
           << " = " << (tree[i].size()) << " (L" << 76 << ")"
           << " "
           << "<stdin>" << endl;
      ;
      int size = 0;
      for (int j = 1; j < tree[i].size(); j++) {
        cerr << "nonleaf[tree[i][j]]"
             << " = " << (nonleaf[tree[i][j]]) << " (L" << 79 << ")"
             << " "
             << "<stdin>" << endl;
        ;
        cerr << "tree[i][j]"
             << " = " << (tree[i][j]) << " (L" << 80 << ")"
             << " "
             << "<stdin>" << endl;
        ;
        if (!nonleaf[tree[i][j]]) size++;
      }
      cerr << "size"
           << " = " << (size) << " (L" << 83 << ")"
           << " "
           << "<stdin>" << endl;
      ;
      if (size < 3) {
        flag = false;
        break;
      }
    }
    if (!flag) break;
  }
  if (flag)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}

### Prompt

Generate a cpp solution 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;
inline int toInt(string s) {
  int v;
  istringstream sin(s);
  sin >> v;
  return v;
}
template <class T>
inline string toString(T x) {
  ostringstream sout;
  sout << x;
  return sout.str();
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
const double EPS = 1e-10;
const double PI = acos(-1.0);
int main(void) {
  vector<vector<int> > tree(1000);
  vector<bool> nonleaf(1000);
  int n;
  cin >> n;
  for (int i = (0); i < (n); ++i) {
    tree[i].push_back(0);
  }
  for (int i = (0); i < (n - 1); ++i) {
    int parent;
    cin >> parent;
    tree[parent - 1].push_back(i + 1);
  }
  for (int i = (0); i < (n - 1); ++i) {
    if (tree[i].size() > 1) {
      cerr << "i"
           << " = " << (i) << endl;
      ;
      nonleaf[i] = true;
    }
  }
  bool flag = true;
  for (int i = (0); i < (n); ++i) {
    if (nonleaf[i]) {
      cerr << "i"
           << " = " << (i) << " (L" << 75 << ")"
           << " "
           << "<stdin>" << endl;
      ;
      cerr << "tree[i].size()"
           << " = " << (tree[i].size()) << " (L" << 76 << ")"
           << " "
           << "<stdin>" << endl;
      ;
      int size = 0;
      for (int j = 1; j < tree[i].size(); j++) {
        cerr << "nonleaf[tree[i][j]]"
             << " = " << (nonleaf[tree[i][j]]) << " (L" << 79 << ")"
             << " "
             << "<stdin>" << endl;
        ;
        cerr << "tree[i][j]"
             << " = " << (tree[i][j]) << " (L" << 80 << ")"
             << " "
             << "<stdin>" << endl;
        ;
        if (!nonleaf[tree[i][j]]) size++;
      }
      cerr << "size"
           << " = " << (size) << " (L" << 83 << ")"
           << " "
           << "<stdin>" << endl;
      ;
      if (size < 3) {
        flag = false;
        break;
      }
    }
    if (!flag) break;
  }
  if (flag)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}

```