#include <bits/stdc++.h>
using namespace std;
struct edge {
  int to, next;
} edges[1111 * 2];
int head[1111];
void addedge(int from, int to, int no) {
  edges[no].to = to;
  edges[no].next = head[from];
  head[from] = no;
}
int ans;
int leaf[1111];
void sol(int o, int fa) {
  int cnt = 0;
  for (int i = head[o]; ~i; i = edges[i].next) {
    int to = edges[i].to;
    if (to != fa) cnt++;
  }
  if (!cnt) {
    leaf[o] = 1;
    return;
  }
  cnt = 0;
  for (int i = head[o]; ~i; i = edges[i].next) {
    int to = edges[i].to;
    if (to == fa) continue;
    sol(to, o);
    if (leaf[to]) cnt++;
  }
  if (cnt < 3) ans = 0;
}
int main() {
  int n;
  cin >> n;
  memset(head, -1, sizeof(head));
  for (int i = 2; i <= n; i++) {
    int a;
    cin >> a;
    addedge(i, a, i << 1);
    addedge(a, i, i << 1 | 1);
  }
  ans = 1;
  sol(1, -1);
  if (ans)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}

### Prompt

In cpp, your task is to solve 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;
struct edge {
  int to, next;
} edges[1111 * 2];
int head[1111];
void addedge(int from, int to, int no) {
  edges[no].to = to;
  edges[no].next = head[from];
  head[from] = no;
}
int ans;
int leaf[1111];
void sol(int o, int fa) {
  int cnt = 0;
  for (int i = head[o]; ~i; i = edges[i].next) {
    int to = edges[i].to;
    if (to != fa) cnt++;
  }
  if (!cnt) {
    leaf[o] = 1;
    return;
  }
  cnt = 0;
  for (int i = head[o]; ~i; i = edges[i].next) {
    int to = edges[i].to;
    if (to == fa) continue;
    sol(to, o);
    if (leaf[to]) cnt++;
  }
  if (cnt < 3) ans = 0;
}
int main() {
  int n;
  cin >> n;
  memset(head, -1, sizeof(head));
  for (int i = 2; i <= n; i++) {
    int a;
    cin >> a;
    addedge(i, a, i << 1);
    addedge(a, i, i << 1 | 1);
  }
  ans = 1;
  sol(1, -1);
  if (ans)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
}

```