#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
void desperate_optimization(bool submitting = true, int precision = 10) {
  if (!submitting) return;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(precision);
}
int main() {
  desperate_optimization(false);
  int n, u, v, x, y, i, j, count = 0, temp, flag1 = 0, flag2 = 0;
  vector<int> tree[1005];
  cin >> n;
  for (i = 2; i <= n; i++) {
    cin >> x;
    tree[x].push_back(i);
  }
  for (i = 1; i <= n; i++) {
    x = tree[i].size();
    if (x) {
      count = 0;
      for (j = 0; j < tree[i].size(); j++) {
        if (tree[tree[i][j]].size() == 0) count++;
      }
      if (count < 3) return cout << "NO", 0;
    }
  }
  return cout << "YES", 0;
  int do_not_hack;
  return 0;
}

### Prompt

Please formulate a cpp solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
void desperate_optimization(bool submitting = true, int precision = 10) {
  if (!submitting) return;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.setf(ios::showpoint);
  cout.precision(precision);
}
int main() {
  desperate_optimization(false);
  int n, u, v, x, y, i, j, count = 0, temp, flag1 = 0, flag2 = 0;
  vector<int> tree[1005];
  cin >> n;
  for (i = 2; i <= n; i++) {
    cin >> x;
    tree[x].push_back(i);
  }
  for (i = 1; i <= n; i++) {
    x = tree[i].size();
    if (x) {
      count = 0;
      for (j = 0; j < tree[i].size(); j++) {
        if (tree[tree[i][j]].size() == 0) count++;
      }
      if (count < 3) return cout << "NO", 0;
    }
  }
  return cout << "YES", 0;
  int do_not_hack;
  return 0;
}

```