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>
using namespace std;
int n, p, cnt, head[1010], vet[1010], Next[1010], size[1010], sum[1010];
inline void add_edge(int u, int v) {
  vet[++cnt] = v;
  Next[cnt] = head[u];
  head[u] = cnt;
}
void dfs(int u) {
  size[u] = 1;
  if (!head[u]) return;
  for (int i = head[u]; i; i = Next[i]) {
    int v = vet[i];
    dfs(v);
    if (size[v] == 1) sum[u]++;
    size[u] += size[v];
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d", &p);
    add_edge(p, i);
  }
  dfs(1);
  for (int i = 1; i <= n; i++)
    if (size[i] > 1)
      if (sum[i] < 3) {
        puts("No");
        return 0;
      }
  puts("Yes");
  return 0;
}