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 PYTHON3):

Solution:
import sys
import math
#import random
#sys.setrecursionlimit(10000000)
input = sys.stdin.readline
 
############ ---- USER DEFINED INPUT FUNCTIONS ---- ############
def inp():
    return(int(input()))
def inara():
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
################################################################
############ ---- THE ACTUAL CODE STARTS BELOW ---- ############

n=inp()

p=[0]*(n+1)
deg=[0]*(n+1)

for i in range(1,n):
	u=inp()
	p[i+1]=u
	deg[u]+=1

for i in range(1,n+1):
	if deg[i]:
		ans=0
		for j in range(1,n+1):
			if p[j]==i and deg[j]==0:
				ans+=1
		if ans<3:
			print("No")
			exit(0)
print("Yes")