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 MAX = 3e5 + 10;
const int RED = 0;
const int BLUE = 1;
int n;
vector<int> g[MAX];
bool is_leaf(int u) { return int(g[u].size() == 0); }
bool dfs(int u, int p) {
  int leaf = 0;
  bool ans = true;
  for (auto v : g[u]) {
    if (is_leaf(v)) {
      leaf++;
    } else {
      ans = ans && dfs(v, u);
    }
  }
  return (ans && leaf >= 3);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int u = 2; u <= n; u++) {
    int v;
    cin >> v;
    g[v].push_back(u);
  }
  cout << (dfs(1, -1) ? "Yes\n" : "NO\n");
  return 0;
}