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;
struct node {
  int parent;
  int no_of_leaf;
  int flag;
  node() {
    no_of_leaf = 0;
    flag = 0;
  }
  int if_leaf() { return flag; }
  void upgrade_leaf() { flag = 1; }
};
int main() {
  int n;
  scanf("%d", &n);
  ;
  node nodes[n + 1];
  nodes[1].upgrade_leaf();
  for (int i = 1; i < n; i++) {
    int par;
    scanf("%d", &par);
    ;
    nodes[i + 1].parent = par;
    nodes[par].no_of_leaf++;
    if (par == 1) continue;
    if (nodes[par].if_leaf() == 0) {
      nodes[par].upgrade_leaf();
      nodes[nodes[par].parent].no_of_leaf--;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (nodes[i].no_of_leaf < 3 && nodes[i].if_leaf() != 0) {
      printf("No\n");
      return 0;
    }
  }
  printf("Yes\n");
  return 0;
}