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;
vector<int> vec[1001];
bool vis[1001];
int k[1001] = {0};
void dfs(int a) {
  vis[a] = true;
  for (int i = 0; i < vec[a].size(); i++) {
    if (!vis[vec[a][i]]) {
      if (vec[vec[a][i]].size() == 0) {
        k[a]++;
      }
      dfs(vec[a][i]);
    }
  }
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int x, y;
  cin >> x;
  for (int i = 1; i < x; i++) {
    cin >> y;
    vec[y].push_back(i + 1);
  }
  dfs(1);
  int r = 0, ans = 0;
  for (int i = 1; i <= x; i++) {
    if (vec[i].size() != 0) {
      r++;
    }
    if (k[i] >= 3) {
      ans++;
    }
  }
  if (r == ans)
    cout << "Yes";
  else
    cout << "No";
}