#include <bits/stdc++.h>
using namespace std;
int dirx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int diry[8] = {1, -1, 0, 0, 1, -1, 1, -1};
string tostring(long long n) {
  if (!n) return "0";
  string ans = "";
  while (n != 0) {
    ans += (n % 10) + '0';
    n /= 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
double toDouble(string s) {
  double ans = 0;
  int i = 0, flag = 0, cnt = 0, ff = 1;
  while (i < s.length()) {
    if (s[i] == '-') {
      ff = -1;
      i++;
      continue;
    }
    if (s[i] == '.') {
      flag = 1;
      i++;
      continue;
    }
    ans *= 10;
    ans += s[i] - '0';
    if (flag) cnt++;
    i++;
  }
  ans *= ff;
  return ans / pow(10, cnt);
}
vector<int> v[1005];
bool solve(int n) {
  queue<int> q;
  q.push(1);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (v[u].empty()) continue;
    int cnt = 0;
    for (int i = 0; i < v[u].size(); i++) {
      int t = v[u][i];
      if (v[t].empty()) cnt++;
      q.push(t);
    }
    if (cnt < 3) return false;
  }
  return true;
}
int main() {
  int n, a;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &a);
    v[a].push_back(i);
  }
  solve(n) ? cout << "Yes" << endl : cout << "No" << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
int dirx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int diry[8] = {1, -1, 0, 0, 1, -1, 1, -1};
string tostring(long long n) {
  if (!n) return "0";
  string ans = "";
  while (n != 0) {
    ans += (n % 10) + '0';
    n /= 10;
  }
  reverse(ans.begin(), ans.end());
  return ans;
}
double toDouble(string s) {
  double ans = 0;
  int i = 0, flag = 0, cnt = 0, ff = 1;
  while (i < s.length()) {
    if (s[i] == '-') {
      ff = -1;
      i++;
      continue;
    }
    if (s[i] == '.') {
      flag = 1;
      i++;
      continue;
    }
    ans *= 10;
    ans += s[i] - '0';
    if (flag) cnt++;
    i++;
  }
  ans *= ff;
  return ans / pow(10, cnt);
}
vector<int> v[1005];
bool solve(int n) {
  queue<int> q;
  q.push(1);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    if (v[u].empty()) continue;
    int cnt = 0;
    for (int i = 0; i < v[u].size(); i++) {
      int t = v[u][i];
      if (v[t].empty()) cnt++;
      q.push(t);
    }
    if (cnt < 3) return false;
  }
  return true;
}
int main() {
  int n, a;
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &a);
    v[a].push_back(i);
  }
  solve(n) ? cout << "Yes" << endl : cout << "No" << endl;
  return 0;
}

```