#include <bits/stdc++.h>
using namespace std;
vector<long long int> adj[100005];
bool visited[100005];
long long int temp = 0;
void dfs(long long int i, long long int parent) {
  if (!visited[i]) {
    visited[i] = true;
    long long int count = 0;
    vector<long long int>::iterator it;
    if (adj[i].size() == 1) {
      if (adj[i][0] != parent) temp = 1;
      return;
    }
    for (it = adj[i].begin(); it != adj[i].end(); it++) {
      if (*it != parent) {
        if (adj[*it].size() == 1) {
          if (adj[*it][0] == i) count += 1;
        }
      }
    }
    if (count < 3) {
      temp = 1;
      return;
    } else {
      for (it = adj[i].begin(); it != adj[i].end(); it++) {
        if (*it != parent) dfs(*it, i);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, i, u, count = 0, j;
  cin >> n;
  vector<long long int>::iterator it;
  memset(visited, false, sizeof(visited));
  visited[1] = true;
  for (i = 2; i <= n; i++) {
    cin >> u;
    adj[i].push_back(u);
    adj[u].push_back(i);
  }
  for (it = adj[1].begin(); it != adj[1].end(); it++) {
    if (adj[*it].size() == 1) {
      if (adj[*it][0] == 1) count += 1;
    }
  }
  if (count < 3)
    cout << "No" << endl;
  else {
    for (it = adj[1].begin(); it != adj[1].end(); it++) dfs(*it, 1);
    if (temp == 1)
      cout << "No" << endl;
    else
      cout << "Yes" << endl;
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<long long int> adj[100005];
bool visited[100005];
long long int temp = 0;
void dfs(long long int i, long long int parent) {
  if (!visited[i]) {
    visited[i] = true;
    long long int count = 0;
    vector<long long int>::iterator it;
    if (adj[i].size() == 1) {
      if (adj[i][0] != parent) temp = 1;
      return;
    }
    for (it = adj[i].begin(); it != adj[i].end(); it++) {
      if (*it != parent) {
        if (adj[*it].size() == 1) {
          if (adj[*it][0] == i) count += 1;
        }
      }
    }
    if (count < 3) {
      temp = 1;
      return;
    } else {
      for (it = adj[i].begin(); it != adj[i].end(); it++) {
        if (*it != parent) dfs(*it, i);
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, i, u, count = 0, j;
  cin >> n;
  vector<long long int>::iterator it;
  memset(visited, false, sizeof(visited));
  visited[1] = true;
  for (i = 2; i <= n; i++) {
    cin >> u;
    adj[i].push_back(u);
    adj[u].push_back(i);
  }
  for (it = adj[1].begin(); it != adj[1].end(); it++) {
    if (adj[*it].size() == 1) {
      if (adj[*it][0] == 1) count += 1;
    }
  }
  if (count < 3)
    cout << "No" << endl;
  else {
    for (it = adj[1].begin(); it != adj[1].end(); it++) dfs(*it, 1);
    if (temp == 1)
      cout << "No" << endl;
    else
      cout << "Yes" << endl;
  }
  return 0;
}

```