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 as sys    
import math as math
#n = int( sys.stdin.readline())

integers = map(int,sys.stdin.readline().split())

# root
n = integers[0]


mn = range(n-1)
for i in range(0,n-1):
	m = map(int,sys.stdin.readline().split())
	mn[i] = m[0]
	
listofzeros = [0] * n

for i in range(0,n-1):
    listofzeros[mn[i]-1]=listofzeros[mn[i]-1]+1
    
ss = 0
for i in range(0,len(listofzeros)):
    if listofzeros[i]>0:
        s = 0
        for j in range(0,n-1):
            if mn[j]==i+1 and listofzeros[j+1] == 0:
                s=s+1
        if s < 3:
            ss = 1

        
if ss ==1:
    print 'NO'
else:
    print 'YES'