#include <bits/stdc++.h>
using namespace std;
const long long inf = 9.2e18;
const int maxn = 2e3 + 70;
const int INF = 2.1e9;
const int maxm = 1e6 + 70;
const int MOD = 1e9 + 7;
const double eps = 1e-7;
const double PI = acos(-1.0);
int n, m, k;
vector<int> G[maxn];
bool leaf[maxn];
void dfs(int u) {
  if (G[u].size() == 0) leaf[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    dfs(G[u][i]);
  }
}
int main() {
  while (cin >> n) {
    memset(leaf, 0, sizeof(leaf));
    for (int i = 0; i <= n; i++) G[i].clear();
    for (int i = 2; i <= n; i++) {
      int p;
      cin >> p;
      G[p].push_back(i);
    }
    dfs(1);
    int flag = 1;
    for (int i = 1; i <= n; i++) {
      if (!leaf[i]) {
        int cnt = 0;
        for (int j = 0; j < G[i].size(); j++)
          if (leaf[G[i][j]]) cnt++;
        if (cnt < 3) {
          flag = 0;
          break;
        }
      }
    }
    if (flag)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
  return 0;
}

### Prompt

Generate 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>
using namespace std;
const long long inf = 9.2e18;
const int maxn = 2e3 + 70;
const int INF = 2.1e9;
const int maxm = 1e6 + 70;
const int MOD = 1e9 + 7;
const double eps = 1e-7;
const double PI = acos(-1.0);
int n, m, k;
vector<int> G[maxn];
bool leaf[maxn];
void dfs(int u) {
  if (G[u].size() == 0) leaf[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    dfs(G[u][i]);
  }
}
int main() {
  while (cin >> n) {
    memset(leaf, 0, sizeof(leaf));
    for (int i = 0; i <= n; i++) G[i].clear();
    for (int i = 2; i <= n; i++) {
      int p;
      cin >> p;
      G[p].push_back(i);
    }
    dfs(1);
    int flag = 1;
    for (int i = 1; i <= n; i++) {
      if (!leaf[i]) {
        int cnt = 0;
        for (int j = 0; j < G[i].size(); j++)
          if (leaf[G[i][j]]) cnt++;
        if (cnt < 3) {
          flag = 0;
          break;
        }
      }
    }
    if (flag)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
  return 0;
}

```