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>
void solve();
int main() {
  while (scanf("%*[ \t\n]*") != EOF) {
    solve();
  }
  return 0;
}
const int MaxN = 1005;
int d[MaxN], n;
std::list<int> vec[MaxN];
void solve() {
  int p;
  scanf("%d", &n);
  memset(d, 0, sizeof(d));
  for (int i = 1; i <= n; i++) vec[i].clear();
  for (int i = 1; i < n; i++) {
    scanf("%d", &p);
    vec[p].push_back(i + 1);
    d[p]++;
  }
  for (int i = 1; i <= n; i++) {
    int cur = 0;
    for (auto v : vec[i]) {
      if (d[v] == 0) cur++;
    }
    if (d[i] && cur < 3) {
      puts("No");
      return;
    }
  }
  puts("Yes");
}