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>
int N;
int degree[1000];
std::vector<std::vector<int>> childs;
bool isSpruce() {
  for (int i = 0; i < N; i++) {
    if (degree[i] == 0) continue;
    int leafs = 0;
    for (const int& v : childs[i])
      if (degree[v] == 0) leafs++;
    if (leafs < 3) return false;
  }
  return true;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  memset(degree, 0, sizeof degree);
  std::cin >> N;
  childs.resize(N);
  int v;
  for (int i = 1; i < N; i++) {
    std::cin >> v;
    --v;
    degree[v]++;
    childs[v].push_back(i);
  }
  if (isSpruce())
    std::cout << "Yes\n";
  else
    std::cout << "No\n";
  return 0;
}