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;
const int N = 1050;
int n;
bool nonl[N];
int tot[N];
int fa[N];
int getint() {
  char ch = getchar();
  int res = 0;
  while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
  bool neg = 0;
  if (ch == '-') {
    neg = 1;
    ch = getchar();
  }
  while ('0' <= ch && ch <= '9') {
    res = res * 10 + ch - '0';
    ch = getchar();
  }
  if (neg) res = -res;
  return res;
}
int main() {
  int i, j;
  n = getint();
  for (i = 2; i <= n; i++) {
    fa[i] = getint();
    nonl[fa[i]] = 1;
  }
  for (i = 2; i <= n; i++) {
    if (!nonl[i]) tot[fa[i]]++;
  }
  for (i = 1; i <= n; i++) {
    if (nonl[i] && tot[i] < 3) break;
  }
  if (i <= n)
    printf("No\n");
  else
    printf("Yes\n");
  return 0;
}