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<vector<int>> adjList;
int N;
bool isLeaf(int node) {
  if (node == 0) return 0;
  if (adjList[node].size() == 0) {
    return true;
  } else {
    return false;
  }
}
int main() {
  scanf("%d", &N);
  adjList.assign(N, vector<int>());
  for (int i = 1; i < N; i++) {
    int par;
    scanf("%d", &par);
    par--;
    adjList[par].push_back(i);
  }
  bool good = true;
  for (int i = 0; i < N; i++) {
    if (!isLeaf(i)) {
      int cnt = 0;
      for (auto adj : adjList[i]) {
        if (isLeaf(adj)) {
          cnt++;
        }
      }
      if (cnt < 3) {
        good = false;
        break;
      }
    }
  }
  if (good) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
  return 0;
}