#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, -1, 1, 1, 1, -1, -1};
int dy[] = {-1, 1, 0, 0, 1, -1, 1, -1};
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int n, m, c = 0;
int cnt;
bool valid(int i) { return i >= 0 && i < n; }
vector<int> v[10000];
bool check() {
  for (int i = 1; i <= n; i++) {
    int c = 0;
    if ((int)v[i].size() != 0) {
      for (int k = 0; k < int(v[i].size()); k++) {
        if ((int)v[v[i][k]].size() == 0) c++;
      }
      if (c < 3) return false;
    }
  }
  return true;
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    v[x].push_back(i);
  }
  if (check())
    cout << "Yes\n";
  else
    cout << "No\n";
}

### 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>
using namespace std;
int dx[] = {0, 0, -1, 1, 1, 1, -1, -1};
int dy[] = {-1, 1, 0, 0, 1, -1, 1, -1};
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
int n, m, c = 0;
int cnt;
bool valid(int i) { return i >= 0 && i < n; }
vector<int> v[10000];
bool check() {
  for (int i = 1; i <= n; i++) {
    int c = 0;
    if ((int)v[i].size() != 0) {
      for (int k = 0; k < int(v[i].size()); k++) {
        if ((int)v[v[i][k]].size() == 0) c++;
      }
      if (c < 3) return false;
    }
  }
  return true;
}
int main() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int x;
    cin >> x;
    v[x].push_back(i);
  }
  if (check())
    cout << "Yes\n";
  else
    cout << "No\n";
}

```