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;
template <class T, class T2>
inline void chkmax(T &x, const T2 &y) {
  if (x < y) x = y;
}
template <class T, class T2>
inline void chkmin(T &x, const T2 &y) {
  if (x > y) x = y;
}
const int MAXN = (1 << 20);
int n;
vector<int> adj[MAXN];
void read() {
  cin >> n;
  for (int i = 2; i <= n; i++) {
    int p;
    cin >> p;
    adj[p].push_back(i);
  }
}
int dfs(int x) {
  if ((int)adj[x].size() == 0) return 1;
  int ret = 0;
  for (int y : adj[x]) ret += dfs(y);
  if (ret < 3) {
    cout << "No" << '\n';
    exit(0);
  }
  return 0;
}
void solve() {
  dfs(1);
  cout << "Yes" << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  read();
  solve();
  return 0;
}