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 PYTHON):

Solution:
import sys
import random
import functools
import re
import itertools
import math
import copy
import bisect
import Queue
from collections import Counter

sys.setrecursionlimit(1000000)

def debug(*_):
    for x in _:
        print >> sys.stderr, x,
    print >> sys.stderr


if __name__ == "__main__":
    n = int(sys.stdin.readline())
    parents = [int(sys.stdin.readline())-1 for i in xrange(n-1)]
    
    is_leaf = [True] * n
    for x in parents:
        is_leaf[x] = False
    
    num_leaf_children = [0] * n # number of leaf children of node i
    
    for (i, x) in enumerate(parents):
        if is_leaf[i+1]:
            num_leaf_children[x] += 1
    
    if all(is_leaf[i] or num_leaf_children[i] >= 3 for i in xrange(n)):
        print "Yes"
    else:
        print "No"