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;
long long n, u, v, num[(long long)(1e5 + 5)], leaf[(long long)(1e5 + 5)];
double ans = 0, cnt = 0;
bool vis[(long long)(1e5 + 5)];
vector<long long> graph[(long long)(1e5 + 5)];
void dfs(long long x, long long parent) {
  if (vis[x]) return;
  vis[x] = 1;
  if (graph[x].size() == 0) {
    num[parent]++;
    leaf[x] = 1;
  }
  for (auto s : graph[x])
    if (!vis[s]) dfs(s, x);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (long long i = 2; i < n + 1; i++) {
    cin >> u;
    graph[u].push_back(i);
  }
  dfs(1, 0);
  for (long long i = 1; i <= n; i++) {
    if (leaf[i] == 0 && num[i] < 3) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES";
  return 0;
}