#include <bits/stdc++.h>
using namespace std;
const long long int MOD = 1e9 + 7;
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long int modinv(long long int a, long long int mod = MOD) {
  long long int x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
long long int power(long long int a, long long int b, long long int m = MOD) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int binomialCoeff(long long int n, long long int k) {
  long long int res = 1;
  if (k > n - k) k = n - k;
  for (long long int i = 0; i < k; ++i) {
    res = (res * (n - i)) % MOD;
    res = (res * modinv(i + 1)) % MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n, x, ans = 0, c = 0, sum = 0;
    cin >> n;
    vector<vector<long long int> > v;
    v.resize(n + 1, vector<long long int>(n + 1));
    set<long long int> st;
    for (long long int(i) = 1; (i) <= (n - 1); ++(i)) {
      cin >> x;
      v[x][i + 1] = 1;
      st.insert(x);
    }
    for (long long int(i) = 1; (i) <= (n); ++(i)) {
      long long int flag = 1;
      c = 0;
      for (long long int(j) = 1; (j) <= (n); ++(j)) {
        if (v[i][j] == 1) flag = 0;
        if (v[i][j] == 1 && st.count(j) == 0) c++;
      }
      if (flag == 0 && c < 3) {
        cout << "No" << endl;
        return 0;
      }
    }
    cout << "Yes" << endl;
  }
}

### 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;
const long long int MOD = 1e9 + 7;
long long int extended_GCD(long long int a, long long int b, long long int &x,
                           long long int &y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long int x1, y1;
  long long int gcd = extended_GCD(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return gcd;
}
long long int modinv(long long int a, long long int mod = MOD) {
  long long int x, y;
  extended_GCD(a, mod, x, y);
  if (x < 0) x += mod;
  return x;
}
long long int power(long long int a, long long int b, long long int m = MOD) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int binomialCoeff(long long int n, long long int k) {
  long long int res = 1;
  if (k > n - k) k = n - k;
  for (long long int i = 0; i < k; ++i) {
    res = (res * (n - i)) % MOD;
    res = (res * modinv(i + 1)) % MOD;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) {
    long long int n, x, ans = 0, c = 0, sum = 0;
    cin >> n;
    vector<vector<long long int> > v;
    v.resize(n + 1, vector<long long int>(n + 1));
    set<long long int> st;
    for (long long int(i) = 1; (i) <= (n - 1); ++(i)) {
      cin >> x;
      v[x][i + 1] = 1;
      st.insert(x);
    }
    for (long long int(i) = 1; (i) <= (n); ++(i)) {
      long long int flag = 1;
      c = 0;
      for (long long int(j) = 1; (j) <= (n); ++(j)) {
        if (v[i][j] == 1) flag = 0;
        if (v[i][j] == 1 && st.count(j) == 0) c++;
      }
      if (flag == 0 && c < 3) {
        cout << "No" << endl;
        return 0;
      }
    }
    cout << "Yes" << endl;
  }
}

```