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 long long MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long A[1001][1001] = {};
  long long n;
  cin >> n;
  for (int i = (0); i < (101); ++i) {
    A[i][0] = 0;
  }
  for (int i = (0); i < (n - 1); ++i) {
    long long t;
    cin >> t;
    A[t][0] += 1;
    A[t][A[t][0]] = i + 2;
  }
  bool m = true;
  long long l = 1;
  while (m && l <= n) {
    long long count = 10;
    if (A[l][0] > 0) {
      count = 0;
      for (int i = (1); i < (A[l][0] + 1); ++i) {
        if (A[(A[l][i])][0] == 0) count++;
      }
    }
    if (count < 3) {
      m = false;
      break;
    }
    l++;
  }
  if (m)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  return 0;
}