#include <bits/stdc++.h>
using namespace std;
vector<int> vg[5000];
int child[20000];
void bfs(int root) {
  int visited[10000];
  memset(visited, 0, sizeof(visited));
  memset(child, 0, sizeof(child));
  visited[root] = 1;
  queue<int> pq;
  pq.push(root);
  while (!pq.empty()) {
    root = pq.front();
    pq.pop();
    for (int i = 0; i < vg[root].size(); i++) {
      int u = vg[root][i];
      if (visited[u] == 0) {
        if (vg[u].size() == 0) child[root]++;
        pq.push(u);
        visited[u] = 1;
      }
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int x;
    cin >> x;
    vg[x].push_back(i + 1);
  }
  bfs(1);
  int ok = 1;
  for (int i = 1; i <= n; i++) {
    if (vg[i].size() > 0) {
      if (child[i] >= 3)
        continue;
      else
        ok = 0;
    }
  }
  ok == 1 ? (cout << "Yes\n") : (cout << "No\n");
  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;
vector<int> vg[5000];
int child[20000];
void bfs(int root) {
  int visited[10000];
  memset(visited, 0, sizeof(visited));
  memset(child, 0, sizeof(child));
  visited[root] = 1;
  queue<int> pq;
  pq.push(root);
  while (!pq.empty()) {
    root = pq.front();
    pq.pop();
    for (int i = 0; i < vg[root].size(); i++) {
      int u = vg[root][i];
      if (visited[u] == 0) {
        if (vg[u].size() == 0) child[root]++;
        pq.push(u);
        visited[u] = 1;
      }
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int x;
    cin >> x;
    vg[x].push_back(i + 1);
  }
  bfs(1);
  int ok = 1;
  for (int i = 1; i <= n; i++) {
    if (vg[i].size() > 0) {
      if (child[i] >= 3)
        continue;
      else
        ok = 0;
    }
  }
  ok == 1 ? (cout << "Yes\n") : (cout << "No\n");
  return 0;
}

```