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;
int outdegree[1050], leafCount[1050];
vector<int> arr[1050];
bool dfs(int node, int par = -1) {
  int size = 0;
  for (int child : arr[node]) {
    if (child != par) {
      if (!dfs(child, node)) {
        return false;
      }
    }
    if (outdegree[child] == 0) {
      size++;
    }
  }
  if (size >= 3 || (size == 0 && arr[node].size() == 0)) {
    return true;
  } else {
    return false;
  }
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, a;
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> a;
    outdegree[a]++;
    arr[a].push_back(i + 1);
  }
  if (dfs(1)) {
    cout << "Yes";
  } else {
    cout << "No";
  }
  return 0;
}