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 a[100009], bbb[1000009], vv[1000000];
map<long long, long long> mp;
map<long long, long long> mp1;
vector<long long> bv[100000];
bool bfs(long long n) {
  queue<long long> q;
  q.push(n);
  while (!q.empty()) {
    long long r = q.front();
    q.pop();
    {
      long long c = 0;
      for (long long i = 0; i < bv[r].size(); i++) {
        long long k = bv[r][i];
        if (bv[k].size() == 0)
          c++;
        else {
          q.push(k);
        }
      }
      if (c < 3) {
        cout << "No";
        ;
        return 0;
      }
    }
  }
  cout << "YES" << endl;
}
int main() {
  cin.sync_with_stdio(false);
  long long i, j, k, l, c, v, b, bvb[1000], aa[1009], bb[1000], n, m, r, t, e,
      f;
  long long x1, x2, y1, y2, x3, x4, x5, x6;
  string s, s1, s2, s3, ss[10000];
  while (cin >> n) {
    for (i = 2; i < n + 1; i++) {
      cin >> m;
      bv[m].push_back(i);
    }
    bfs(1);
  }
}