#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long power(long long a, long long b, long long modi) {
  a %= modi;
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % modi;
    }
    b /= 2;
    a = (a * a) % modi;
  }
  return res;
}
vector<int> vec[1001];
bool leaf[1001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 2; i <= n; i++) {
      int x;
      cin >> x;
      vec[x].push_back(i);
    }
    for (int i = 2; i <= n; i++) {
      if (vec[i].empty()) leaf[i] = 1;
    }
    bool f = true;
    for (int i = 1; i <= n; i++) {
      if (vec[i].empty()) continue;
      int c = 0;
      for (int j = 0; j < (int)vec[i].size(); j++) {
        if (leaf[vec[i][j]]) c++;
      }
      if (c < 3) {
        f = false;
        break;
      }
    }
    if (f)
      cout << "Yes";
    else
      cout << "No";
  }
  return 0;
}

### Prompt

Your task is to create 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,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long power(long long a, long long b, long long modi) {
  a %= modi;
  long long res = 1;
  while (b) {
    if (b % 2) {
      res = (res * a) % modi;
    }
    b /= 2;
    a = (a * a) % modi;
  }
  return res;
}
vector<int> vec[1001];
bool leaf[1001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 2; i <= n; i++) {
      int x;
      cin >> x;
      vec[x].push_back(i);
    }
    for (int i = 2; i <= n; i++) {
      if (vec[i].empty()) leaf[i] = 1;
    }
    bool f = true;
    for (int i = 1; i <= n; i++) {
      if (vec[i].empty()) continue;
      int c = 0;
      for (int j = 0; j < (int)vec[i].size(); j++) {
        if (leaf[vec[i][j]]) c++;
      }
      if (c < 3) {
        f = false;
        break;
      }
    }
    if (f)
      cout << "Yes";
    else
      cout << "No";
  }
  return 0;
}

```