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>
#pragma GCC optimize("O3")
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long llinf = LLONG_MAX;
const int inf = INT_MAX;
const int nmax = 1e5 + 5;
const int mod = 1e9 + 7;
using namespace std;
int n, i, x, ok;
vector<int> a[1001];
void dfs(int x) {
  int v = 0;
  for (int i = 0; i < a[x].size(); i++) {
    if (a[a[x][i]].empty()) v++;
    dfs(a[x][i]);
  }
  if (v < 3 && !a[x].empty()) ok = 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cerr.tie(0);
  cout.tie(0);
  cin >> n;
  for (i = 1; i < n; i++) {
    cin >> x;
    a[x].push_back(i + 1);
  }
  dfs(1);
  if (ok)
    cout << "No" << endl;
  else
    cout << "Yes" << endl;
  return 0;
}