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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << "\n";
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
inline int sbt(int x) { return __builtin_popcount(x); }
inline long long gl(long long a, long long b) { return (b - a + 1); }
inline long long mul(long long a, long long b,
                     long long m = (long long)(1e9 + 7)) {
  return ((a % m) * (b % m)) % m;
}
inline long long add(long long a, long long b,
                     long long m = (long long)(1e9 + 7)) {
  return (a + b) % m;
}
inline long long sub(long long a, long long b,
                     long long m = (long long)(1e9 + 7)) {
  return (a - b + m) % m;
}
long long fastpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
long long modinv(long long a) { return fastpow(a, (long long)(1e9 + 7) - 2); }
vector<long long> g[1010];
long long dfs(long long u) {
  if (g[u].empty()) return true;
  bool ans = true;
  int c = 0;
  for (auto i : g[u]) {
    if (g[i].empty())
      c++;
    else
      ans = ans and dfs(i);
  }
  if (c > 2) return ans;
  return false;
}
void get_ac() {
  long long n, a;
  cin >> n;
  for (auto i = 2; i <= n; i++) {
    cin >> a, g[a].push_back(i);
  }
  if (dfs(1)) {
    cout << "Yes";
  } else {
    cout << "No";
  }
}
int main() {
  cin.sync_with_stdio(false);
  cout.sync_with_stdio(false);
  cin.tie(NULL);
  { get_ac(); }
  return 0;
}