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;
typedef struct graph {
  int v;
  list<int> *adj;
} graph;
graph *creategraph() {
  graph *gra = new graph;
  cin >> gra->v;
  gra->v++;
  gra->adj = new list<int>[gra->v];
  return gra;
}
void addedge(graph *gra, int u, int v) { gra->adj[u].push_back(v); }
bool bfs(graph *gra) {
  list<int> queue;
  queue.push_back(1);
  int s;
  list<int>::iterator itr;
  while (!queue.empty()) {
    int count = 0;
    s = queue.front();
    queue.pop_front();
    for (itr = gra->adj[s].begin(); itr != gra->adj[s].end(); itr++) {
      if (gra->adj[*itr].empty()) {
        count++;
        continue;
      } else
        queue.push_back(*itr);
    }
    if (count < 3) return false;
  }
  return true;
}
int main() {
  graph *gra = creategraph();
  int i, u;
  for (i = 2; i < gra->v; i++) {
    cin >> u;
    addedge(gra, u, i);
  }
  if (bfs(gra))
    cout << "Yes";
  else
    cout << "No";
}