#include <bits/stdc++.h>
const long long LINF = 1e18;
const int INF = 1e9;
const int M = 1e9 + 7;
const double EPS = 1e-9;
using namespace std;
vector<vector<int> > g;
vector<int> isnode(100000);
vector<int> ch(100000);
vector<int> p(100000);
bool used[10000];
void dfs(int v) {
  used[v] = 1;
  if (g[v].size() == 0) {
    isnode[v] = 1;
  }
  for (int i = 0; i < (int)(g[v].size()); i++) {
    if (!used[g[v][i]]) dfs(g[v][i]);
  }
}
int main(int argc, const char* argv[]) {
  int n;
  cin >> n;
  g.resize(n);
  for (int i = 0; i < (int)(n - 1); i++) {
    int tmp;
    cin >> tmp;
    g[--tmp].push_back(i + 1);
    p[i + 1] = tmp;
  }
  dfs(0);
  for (int i = 0; i < (int)(n); i++) {
    int flag = 0;
    for (int j = 0; j < (int)(g[i].size()); j++) {
      if (isnode[g[i][j]] == 1) flag++;
    }
    if (flag < 3 && isnode[i] == 0) {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes";
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
const long long LINF = 1e18;
const int INF = 1e9;
const int M = 1e9 + 7;
const double EPS = 1e-9;
using namespace std;
vector<vector<int> > g;
vector<int> isnode(100000);
vector<int> ch(100000);
vector<int> p(100000);
bool used[10000];
void dfs(int v) {
  used[v] = 1;
  if (g[v].size() == 0) {
    isnode[v] = 1;
  }
  for (int i = 0; i < (int)(g[v].size()); i++) {
    if (!used[g[v][i]]) dfs(g[v][i]);
  }
}
int main(int argc, const char* argv[]) {
  int n;
  cin >> n;
  g.resize(n);
  for (int i = 0; i < (int)(n - 1); i++) {
    int tmp;
    cin >> tmp;
    g[--tmp].push_back(i + 1);
    p[i + 1] = tmp;
  }
  dfs(0);
  for (int i = 0; i < (int)(n); i++) {
    int flag = 0;
    for (int j = 0; j < (int)(g[i].size()); j++) {
      if (isnode[g[i][j]] == 1) flag++;
    }
    if (flag < 3 && isnode[i] == 0) {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes";
  return 0;
}

```