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 T>
void print(T t) {
  cout << t << endl;
}
long long n = 0, t = 0;
long long small = -1e18, big = 1e18;
const long long mod = 1000000007;
const long long N = 2e5;
map<long long, long long> occ;
map<pair<long long, long long>, long long> occP;
long long factorial(int n) {
  long long ans = 1;
  for (int i = 1; i <= n; ++i) {
    ans *= i;
  }
  return ans;
}
long long findGCD(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= min(a, b); ++i) {
    if (a % i == 0 && b % i == 0) {
      ans = i;
    }
  }
  return ans;
}
bool sortVec(const vector<long long>& v1, const vector<long long>& v2) {
  return v1[0] < v2[0];
}
bool isprime(long long p) {
  for (int i = 2; i < pow(p, 1.5); ++i) {
    if (p % i == 0 && p != i) {
      return false;
    }
  }
  return true;
}
void print2dVec(vector<vector<long long>> v) {
  for (int i = (0); i < (v.size()); ++i) {
    cout << i << " : ";
    for (int j = (0); j < (v[i].size()); ++j) {
      cout << v[i][j] << " ";
    }
    print("");
  }
}
vector<vector<long long>> path(N);
bool ans = true;
void dfs(int curr, int par) {
  int leaf_child = 0;
  for (int child : path[curr]) {
    if (child == par) continue;
    if (path[child].size() == 1) {
      leaf_child += 1;
    } else {
      dfs(child, curr);
    }
  }
  if (leaf_child < 3) {
    ans = false;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  cin >> n;
  for (int i = 2; i <= n; ++i) {
    long long par;
    cin >> par;
    path[par].push_back(i);
    path[i].push_back(par);
  }
  dfs(1, -1);
  if (ans) {
    print("Yes");
  } else {
    print("No");
  }
  return 0;
}