#include <bits/stdc++.h>
int n;
int edgenum;
int vet[10000];
int nextx[10000];
int head[10000];
int vis[10000];
int size[10000];
void add(int u, int v) {
  edgenum++;
  vet[edgenum] = v;
  nextx[edgenum] = head[u];
  head[u] = edgenum;
}
bool dfs(int u) {
  vis[u] = true;
  int cnt = 0;
  size[u] = 1;
  for (int i = head[u]; i; i = nextx[i]) {
    int v = vet[i];
    if (vis[v]) {
      return false;
    }
    if (!dfs(v)) {
      return false;
    }
    size[u] += size[v];
    if (size[v] == 1) {
      cnt++;
    }
  }
  if (size[u] != 1 && cnt < 3) {
    return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int father;
    scanf("%d", &father);
    add(father, i);
  }
  if (dfs(1)) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
  return 0;
}

### Prompt

Generate 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>
int n;
int edgenum;
int vet[10000];
int nextx[10000];
int head[10000];
int vis[10000];
int size[10000];
void add(int u, int v) {
  edgenum++;
  vet[edgenum] = v;
  nextx[edgenum] = head[u];
  head[u] = edgenum;
}
bool dfs(int u) {
  vis[u] = true;
  int cnt = 0;
  size[u] = 1;
  for (int i = head[u]; i; i = nextx[i]) {
    int v = vet[i];
    if (vis[v]) {
      return false;
    }
    if (!dfs(v)) {
      return false;
    }
    size[u] += size[v];
    if (size[v] == 1) {
      cnt++;
    }
  }
  if (size[u] != 1 && cnt < 3) {
    return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    int father;
    scanf("%d", &father);
    add(father, i);
  }
  if (dfs(1)) {
    printf("Yes\n");
  } else {
    printf("No\n");
  }
  return 0;
}

```