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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000;
vector<int> adj[maxn];
bool pos = true;
void dfs(int n, int parent) {
  int cur = 0;
  for (int i : adj[n]) {
    if (i == parent) {
      continue;
    }
    if (adj[i].size() == 1) {
      ++cur;
    } else {
      dfs(i, n);
      if (!pos) {
        return;
      }
    }
  }
  if (cur < 3) {
    pos = false;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, a;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    cin >> a;
    --a;
    adj[a].push_back(i + 1);
    adj[i + 1].push_back(a);
  }
  dfs(0, -1);
  if (pos) {
    cout << "Yes"
         << "\n";
  } else {
    cout << "No"
         << "\n";
  }
  return 0;
}