#include <bits/stdc++.h>
using namespace std;
vector<int> graph[1008];
int visited[1008];
int degree[1008];
int children[1008];
int main() {
  int n, a;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> a;
    degree[a]++;
    graph[a].push_back(i);
  }
  for (int i = 0; i < 1008; i++) {
    visited[i] = 0;
    children[i] = 0;
  }
  queue<int> q;
  q.push(1);
  visited[1] = 1;
  while (!q.empty()) {
    int x = q.front();
    for (int i = 0; i < graph[x].size(); i++) {
      int now = graph[x][i];
      if (!visited[now]) {
        visited[now] = true;
        if (degree[now] == 0) {
          children[x]++;
        }
        q.push(now);
      }
    }
    if (children[x] < 3 && degree[x] != 0) {
      cout << "No" << endl;
      return 0;
    }
    q.pop();
  }
  cout << "Yes" << endl;
}

### Prompt

Your task is to create 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<int> graph[1008];
int visited[1008];
int degree[1008];
int children[1008];
int main() {
  int n, a;
  cin >> n;
  for (int i = 2; i <= n; i++) {
    cin >> a;
    degree[a]++;
    graph[a].push_back(i);
  }
  for (int i = 0; i < 1008; i++) {
    visited[i] = 0;
    children[i] = 0;
  }
  queue<int> q;
  q.push(1);
  visited[1] = 1;
  while (!q.empty()) {
    int x = q.front();
    for (int i = 0; i < graph[x].size(); i++) {
      int now = graph[x][i];
      if (!visited[now]) {
        visited[now] = true;
        if (degree[now] == 0) {
          children[x]++;
        }
        q.push(now);
      }
    }
    if (children[x] < 3 && degree[x] != 0) {
      cout << "No" << endl;
      return 0;
    }
    q.pop();
  }
  cout << "Yes" << endl;
}

```