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 = 2e5 + 3;
const int M = 1e7 + 7;
const long long MOD = 1e9 + 7;
const long long INF = 1e18 + 9;
const int SM = 1e3 + 5;
const int logN = 20, BL = 400;
const double EPS = 3e-16;
const int dx[] = {1, 0, 0, -1, -1, 1, -1, 1};
const int dy[] = {0, 1, -1, 0, -1, 1, 1, -1};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> p(n + 1), deg(n + 1), v(n + 1);
  for (long long i = 2; i <= n; i++) {
    cin >> p[i];
    deg[p[i]]++;
  }
  for (long long i = 1; i <= n; i++) {
    if (!deg[i]) {
      v[p[i]]++;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (deg[i] > 0 && v[i] < 3) return cout << "No", 0;
  }
  cout << "Yes";
  return 0;
}