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:
n=int(input())
arr = [0]*(n+2)
arri = [False]*(n+2)
itrmap = dict()
mitrmap = dict()
# root = 1
# previnp = 0
for i in range(n-1):
	inp = int(input())
	# if previnp!=inp:
	# 	itr = i+2
	# 	previnp = inp
	# arr[inp]+=1
	# itrmap[inp]=[(itr,i+2)]
	# maplist = list(itrmap.keys())
	# #maplist.sort()
	# if inp>=itrmap[root][0] and inp<=itrmap[root][1]:
	# 	pass
	# else:
	# 	for j in maplist:
	# 		if inp>=itrmap[j][0] and inp<=itrmap[j][1]:
	# 			root = j
	# 			break
	# if inp>=itrmap[root][0] and inp<=itrmap[root][1] and arri[inp]==False:
	# 	arr[root]-=1
	# 	arri[inp]=True
# 	print(maplist,root)
# print(arr)
# print(itrmap)
# flag = 0
# for j in maplist:
# 	if arr[j]<3:
# 		flag = 1
# 		break

# if flag == 1:
# 	print("No")
# else:
# 	print("Yes")
	arr[i+2]=inp
	try:
		itrmap[inp].append(i+2)
		mitrmap[inp]+=1
	except:
		itrmap[inp] = [i+2]
		mitrmap[inp] = 1
	try:
		# pass
		if arri[inp]==False:
			mitrmap[arr[inp]]-=1
			arri[inp]=True
	except:
		pass

# print(arr)
# print(itrmap)
# print(mitrmap)
maplist = list(mitrmap.keys())
flag = 0
for j in maplist:
	if (mitrmap[j])<3:
		flag = 1
		break
if flag == 1:
	print("No")
else:
	print("Yes")