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 INF = 1 << 30;
const int MAX = 1010;
const int mod = 1e9 + 7;
vector<int> ind[MAX];
int num[MAX];
int t[MAX];
int parent[MAX];
bool v[MAX];
string s = "Yes";
void search(int i) {
  if (ind[i].size() > 0) {
    num[parent[i]]--;
    for (int j = 0; j < ind[i].size(); j++) {
      search(ind[i][j]);
    }
  } else {
    v[i] = true;
  }
}
int main(int argc, char *argv[]) {
  int i, j, k, N;
  cin >> N;
  for (i = 1; i < N; i++) {
    scanf("%d", &t[i]);
  }
  for (i = 1; i < N; i++) {
    ind[t[i]].push_back(i + 1);
    num[t[i]]++;
    parent[i + 1] = t[i];
  }
  search(1);
  for (i = 1; i < N; i++) {
    if (!v[i] && num[i] < 3) {
      s = "No";
      break;
    }
  }
  cout << s << endl;
  return 0;
}