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;
static int N, p;
static vector<int> child[1010];
static bool flag = true;
void check(int x) {
  if (child[x].size() == 0 || flag == false) {
    return;
  }
  int cnt = 0;
  for (int i = 0; i < child[x].size(); i++) {
    int v = child[x][i];
    if (child[v].size() == 0)
      ++cnt;
    else
      check(v);
  }
  if (cnt < 3 && child[x].size() > 0) flag = false;
}
int main() {
  while (scanf("%d", &N) != EOF) {
    for (int i = 1; i <= N; i++) child[i].clear();
    for (int i = 1; i <= N - 1; i++) {
      scanf("%d", &p);
      child[p].push_back(i + 1);
    }
    flag = true;
    check(1);
    printf(flag == 1 ? "Yes\n" : "No\n");
  }
  return 0;
}