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 $*b*_{1}, *b*_{2}, ..., *b*_{*n*}$, consisting of zeros and ones. Each second Pavel move skewer on position $*i*$ to position $*p*_{*i*}$, and if $*b*_{*i*}$ 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 $2*n*$ 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 $*k*$ ($*k* ≥ 2*n*$), so that after $*k*$ seconds each skewer visits each of the $2*n*$ 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 $*n*$ ($1 ≤ *n* ≤ 2·10^{5}$) — the number of skewers.

The second line contains a sequence of integers $*p*_{1}, *p*_{2}, ..., *p*_{*n*}$ ($1 ≤ *p*_{*i*} ≤ *n*$) — the permutation, according to which Pavel wants to move the skewers.

The third line contains a sequence $*b*_{1}, *b*_{2}, ..., *b*_{*n*}$ 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 $2*n*$ placements.

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 $1$.