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 N = 1e3 + 5;
const long long INF = 1ll * 0x3f3f3f3f;
bool avail[N], check = true;
vector<int> adj[N];
void DFS(int u) {
  if (check == false) return;
  int tmp = adj[u].size();
  if (adj[u].size() > 0) avail[u] = false;
  for (int i = (0); i < (adj[u].size()); ++i) {
    int v = adj[u][i];
    DFS(v);
    if (avail[v] == false) tmp--;
  }
  if (avail[u] == false && tmp < 3) {
    check = false;
    return;
  }
}
int main() {
  int n;
  cin >> n;
  memset(avail, true, sizeof avail);
  for (int i = (2); i <= (n); ++i) {
    int p;
    cin >> p;
    adj[p].push_back(i);
  }
  DFS(1);
  if (check == true)
    cout << "Yes";
  else
    cout << "No";
  return 0;
}