(Analysis by Lewin Gan)

Let's consider the problem without any cyclic shifts.

Let $a_i$ be the position of the $i$th cow in the first line, and $b_i$ be the position of the $i$th cow in the second sequence.

Two cows $i,j$ will cross if $(a_i < a_j) \neq (b_i < b_j)$.

Define a sequence $s$ such that $s_{a_i} = b_i$. Intuitively, this encodes the position of the $k$th cow in the first line from the left in the second line.

So, we can rewrite our condition of two cows intersecting as $p < q$ and $s_p > s_q$. Note this looks like the formula for number of inversions.

So, initially we can get the number of crossings without any cyclic shifts by counting the number of inversions. This is a classic problem that can be done with divide and conquer or with a binary indexed tree.

Now, let's consider what happens when we move one cow from the end of the line to the beginning. Suppose this cow ends up in position $P$ in the second line. Since this cow is the last cow in the first line, we can see it contributes $N-P$ intersections. Moving it to the beginning would then contribute $P-1$ intersections. So moving it from the first position to last position would change the number of pairs by $(P-1) - (N-P)$.

Below is Mark Chen's implementation. Note that he uses divide and conquer to count the number of intersections at the beginning.

#include <iostream>
#include <iomanip>
#include <stdio.h>
#include <set>
#include <vector>
#include <map>
#include <cmath>
#include <algorithm>
#include <memory.h>
#include <string>
#include <sstream>
#include <cstdlib>
#include <ctime>
#include <cassert>

using namespace std;

typedef long long LL;
typedef pair<int,int> PII;

#define MP make_pair
#define PB push_back
#define FF first
#define SS second

#define FORN(i, n) for (int i = 0; i <  (int)(n); i++)
#define FOR1(i, n) for (int i = 1; i <= (int)(n); i++)
#define FORD(i, n) for (int i = (int)(n) - 1; i >= 0; i--)

#define DEBUG(X) { cout << #X << " = " << (X) << endl; }
#define PR0(A,n) { cout << #A << " = "; FORN(_,n) cout << A[_] << ' '; cout << endl; }

#define MOD 1000000007
#define INF 2000000000

int GLL(LL& x) {
return scanf("%lld", &x);
}

int GI(int& x) {
return scanf("%d", &x);
}

const int MAXN = 100005;
int q[2][MAXN], qInv[2][MAXN], qCopy[2][MAXN];

int n, x;

vector<int> merged;

LL inversions(int p[], int l, int r) {
if (l != r) {
int m = (l+r) / 2;  // [l,m], [m+1,r]

LL res = inversions(p, l, m) + inversions(p, m + 1, r);

merged.clear();

int i = l; int j = m + 1;

while (i <= m && j <= r) {
if (p[i] < p[j]) {
res += j - (m + 1);
merged.PB(p[i]);
i++;
}
else {
merged.PB(p[j]);
j++;
}
}
while (i <= m) {
res += j - (m + 1);
merged.PB(p[i]);
i++;
}
while (j <= r) {
merged.PB(p[j]);
j++;
}

i = 0;
for (j = l; j <= r; j++) {
p[j] = merged[i];
i++;
}

return res;
}
else {
return 0;
}
}

int main() {
GI(n);

FORN(seq, 2) FOR1(i, n) {
GI(q[seq][i]);
qInv[seq][q[seq][i]] = i;
}

FORN(seq, 2) FOR1(i, n) {
qCopy[seq][i] = qInv[1-seq][q[seq][i]];
}

FORN(seq, 2) FOR1(i, n) {
q[seq][i] = qCopy[seq][i];
}

LL res = 1LL*INF*INF;

FORN(seq, 2) {
LL cur = inversions(q[seq], 1, n);

for (int i = n; i >= 1; i--) {
cur += 2 * qCopy[seq][i] - n - 1;
res = min(res, cur);
}
}

printf("%lld\n", res);
return 0;
}