Pavel cooks barbecue. There are n skewers, they lay on a brazier in a row, each on one of n positions. Pavel wants each skewer to be cooked some time in every of n positions in two directions: in the one it was directed originally and in the reversed direction.

Pavel has a plan: a permutation p and a sequence b1, b2, ..., bn, consisting of zeros and ones. Each second Pavel move skewer on position i to position pi, and if bi equals 1 then he reverses it. So he hope that every skewer will visit every position in both directions.

Unfortunately, not every pair of permutation p and sequence b suits Pavel. What is the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements? Note that after changing the permutation should remain a permutation as well.

There is no problem for Pavel, if some skewer visits some of the placements several times before he ends to cook. In other words, a permutation p and a sequence b suit him if there is an integer k (k ≥ 2n), so that after k seconds each skewer visits each of the 2n placements.

It can be shown that some suitable pair of permutation p and sequence b exists for any n.

Input

The first line contain the integer n (1 ≤ n ≤ 2·105) — the number of skewers.

The second line contains a sequence of integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the permutation, according to which Pavel wants to move the skewers.

The third line contains a sequence b1, b2, ..., bn consisting of zeros and ones, according to which Pavel wants to reverse the skewers.

Output

Print single integer — the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements.

Examples

Input

4
4 3 2 1
0 1 1 1


Output

2


Input

3
2 3 1
0 0 0


Output

1

Note

In the first example Pavel can change the permutation to 4, 3, 1, 2.

In the second example Pavel can change any element of b to 1.
class Graph:
	def __init__(self,n):	
		self.nodes=n
		self.edge_mapping=[]
		for x in range(self.nodes+1):
			self.edge_mapping.append([])
	def add_edge(self,src,dest):
		if self.edge_mapping[src]==None:
			self.edge_mapping[src]=[dest]
		else:
			self.edge_mapping[src].append(dest)
	
	def dfs_inner(self,nd,mark_list):
		stack=[]
		stack.append(nd)
		while len(stack)!=0:
			node=stack.pop()
			mark_list[node]=1
			for n in self.edge_mapping[node]:
				if mark_list[n]==0:
					stack.append(n)
		
	def dfs(self):
		mark_list=[0]*(self.nodes+1)
		ret=0;
		for i in range(1,self.nodes+1):
			if mark_list[i]!=1:
				self.dfs_inner(i,mark_list)
				ret+=1
		return ret

	def printg(self):
		for z in self.edge_mapping:
			print(z)
	

if __name__ == '__main__':
	n=int(input())
	g=Graph(n)
	p_list=list(map(int,input().split()))
	s_list=list(map(int,input().split()))
	for x in range(len(p_list)):
		g.add_edge(p_list[x],x+1)
	ans=0
	trav=g.dfs()
	if trav!=1:
		ans+=trav
	s=0
	for x in s_list:
		s+=x
	# print(ans)
	if s%2==0:
		print(ans+1)
	else:
		print(ans)