#include <bits/stdc++.h>
using namespace std;
long long cnt, nb;
void SieveOfErat() {
  bool prime[1000001];
  memset(prime, true, sizeof(prime));
  for (long long ind = 2; ind * ind <= 1000000; ind++) {
    if (prime[ind] == true) {
      for (long long cd = ind * 2; cd <= 1000000; cd += ind) prime[cd] = false;
    }
  }
}
long long gcd(long long st, long long y) {
  if (st == 0) {
    return y;
  } else
    return gcd(st % y, y);
}
struct vect {
  struct vect *foll;
  long long val;
  struct vect *succ;
};
void upd(struct vect **beg, struct vect **end, long long xd) {
  struct vect *c = (struct vect *)malloc(sizeof(struct vect));
  c->val = xd;
  c->succ = NULL;
  if (*beg == NULL) {
    c->foll = NULL;
    *beg = c;
    *end = c;
    return;
  }
  (*end)->succ = c;
  c->foll = *end;
  *end = c;
}
signed main() {
  long long dr, orf;
  scanf("%lld", &dr);
  vector<long long> v(dr + 1), p(dr + 1);
  vector<vector<long long> > mat(dr + 1);
  for (long long cd = 2; cd <= dr; cd++) {
    scanf("%lld", &orf);
    mat[orf].push_back(cd);
  }
  for (long long cd = 1; cd <= dr; cd++) {
    if (mat[cd].size() == 0) continue;
    long long cnt = 0;
    for (auto &xc : mat[cd]) {
      if (mat[xc].size() == 0) cnt++;
    }
    if (cnt < 3) {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}

### Prompt

Please formulate 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;
long long cnt, nb;
void SieveOfErat() {
  bool prime[1000001];
  memset(prime, true, sizeof(prime));
  for (long long ind = 2; ind * ind <= 1000000; ind++) {
    if (prime[ind] == true) {
      for (long long cd = ind * 2; cd <= 1000000; cd += ind) prime[cd] = false;
    }
  }
}
long long gcd(long long st, long long y) {
  if (st == 0) {
    return y;
  } else
    return gcd(st % y, y);
}
struct vect {
  struct vect *foll;
  long long val;
  struct vect *succ;
};
void upd(struct vect **beg, struct vect **end, long long xd) {
  struct vect *c = (struct vect *)malloc(sizeof(struct vect));
  c->val = xd;
  c->succ = NULL;
  if (*beg == NULL) {
    c->foll = NULL;
    *beg = c;
    *end = c;
    return;
  }
  (*end)->succ = c;
  c->foll = *end;
  *end = c;
}
signed main() {
  long long dr, orf;
  scanf("%lld", &dr);
  vector<long long> v(dr + 1), p(dr + 1);
  vector<vector<long long> > mat(dr + 1);
  for (long long cd = 2; cd <= dr; cd++) {
    scanf("%lld", &orf);
    mat[orf].push_back(cd);
  }
  for (long long cd = 1; cd <= dr; cd++) {
    if (mat[cd].size() == 0) continue;
    long long cnt = 0;
    for (auto &xc : mat[cd]) {
      if (mat[xc].size() == 0) cnt++;
    }
    if (cnt < 3) {
      cout << "No";
      return 0;
    }
  }
  cout << "Yes" << endl;
  return 0;
}

```