#include <bits/stdc++.h>
using namespace std;
int arr[1005], areleaf[1005];
bool vis[1005];
vector<int> v[1005];
bool poss = true;
int dfs(int u) {
  int leaf = 0;
  bool flag = false;
  vis[u] = true;
  for (int i = 0; i < v[u].size(); i++) {
    if (!vis[v[u][i]]) {
      flag = true;
      leaf += dfs(v[u][i]);
    }
  }
  if (!flag) {
    areleaf[u] = 1;
    return 1;
  } else {
    arr[u] = leaf;
    return 0;
  }
}
int main() {
  int n, x;
  (scanf("%d", &n));
  for (int i = 1; i <= n - 1; i++) {
    (scanf("%d", &x));
    v[x].push_back(i + 1);
    v[i + 1].push_back(x);
  }
  memset(vis, false, sizeof(vis));
  memset(areleaf, 0, sizeof(areleaf));
  dfs(1);
  for (int i = 1; i <= n; i++) {
    if (areleaf[i] == 0) {
      if (arr[i] < 3) {
        cout << "No";
        return 0;
      }
    }
  }
  cout << "Yes";
}

### Prompt

Please formulate 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;
int arr[1005], areleaf[1005];
bool vis[1005];
vector<int> v[1005];
bool poss = true;
int dfs(int u) {
  int leaf = 0;
  bool flag = false;
  vis[u] = true;
  for (int i = 0; i < v[u].size(); i++) {
    if (!vis[v[u][i]]) {
      flag = true;
      leaf += dfs(v[u][i]);
    }
  }
  if (!flag) {
    areleaf[u] = 1;
    return 1;
  } else {
    arr[u] = leaf;
    return 0;
  }
}
int main() {
  int n, x;
  (scanf("%d", &n));
  for (int i = 1; i <= n - 1; i++) {
    (scanf("%d", &x));
    v[x].push_back(i + 1);
    v[i + 1].push_back(x);
  }
  memset(vis, false, sizeof(vis));
  memset(areleaf, 0, sizeof(areleaf));
  dfs(1);
  for (int i = 1; i <= n; i++) {
    if (areleaf[i] == 0) {
      if (arr[i] < 3) {
        cout << "No";
        return 0;
      }
    }
  }
  cout << "Yes";
}

```