#include <bits/stdc++.h>
#pragma comment(linker, "/stack:20000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const int inf = (int)2e9, mod = (int)1e9 + 7;
const long long linf = (long long)2e18;
const int maxn = (int)5e5 + 1;
const int maxn2 = (int)1e6 + 1;
void precalc() {}
void clear() {}
int n;
vector<int> g[maxn];
void read() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int p;
    cin >> p;
    g[p].push_back(i + 1);
  }
}
bool used[maxn];
void dfs(int x) {
  if (!(int)g[x].size()) return;
  used[x] = true;
  int cnt = 0;
  for (int i = 0; i < (int)g[x].size(); ++i) {
    if (!(int)g[g[x][i]].size()) ++cnt;
    if (!used[g[x][i]]) dfs(g[x][i]);
  }
  if (cnt < 3) cout << "No", exit(0);
  return;
}
void solve() {
  read();
  dfs(1);
  cout << "Yes";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  precalc();
  int t = 1;
  if (0) cin >> t;
  for (int i = 1; i <= t; ++i) {
    if (0) cout << "Case " << i << ": ";
    solve();
    if (i != t) clear();
  }
  exit(0);
}

### Prompt

Your challenge is to write 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>
#pragma comment(linker, "/stack:20000000")
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
using namespace std;
const int inf = (int)2e9, mod = (int)1e9 + 7;
const long long linf = (long long)2e18;
const int maxn = (int)5e5 + 1;
const int maxn2 = (int)1e6 + 1;
void precalc() {}
void clear() {}
int n;
vector<int> g[maxn];
void read() {
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int p;
    cin >> p;
    g[p].push_back(i + 1);
  }
}
bool used[maxn];
void dfs(int x) {
  if (!(int)g[x].size()) return;
  used[x] = true;
  int cnt = 0;
  for (int i = 0; i < (int)g[x].size(); ++i) {
    if (!(int)g[g[x][i]].size()) ++cnt;
    if (!used[g[x][i]]) dfs(g[x][i]);
  }
  if (cnt < 3) cout << "No", exit(0);
  return;
}
void solve() {
  read();
  dfs(1);
  cout << "Yes";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  precalc();
  int t = 1;
  if (0) cin >> t;
  for (int i = 1; i <= t; ++i) {
    if (0) cout << "Case " << i << ": ";
    solve();
    if (i != t) clear();
  }
  exit(0);
}

```