#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vll = vector <ll>;
using ii = pair <ll, ll>;
using vii = vector <ii>;
ll freq[100005][3];
// ii operator+ (const ii a, const ii b) { return ii{ a.first + b.first, a.second + b.second, }; }
// ii operator- (const ii a, const ii b) { return ii{ a.first - b.first, a.second - b.second, }; }
// ii operator+= (ii &a, const ii b) { return a = a + b; }
// ostream& operator<< (ostream &os, const ii a) { return os << a.first << ' ' << a.second; }
int main () {
cin.tie(nullptr) -> sync_with_stdio(false);
ll n;
cin >> n;
vii ve(2*n);
for (auto &[x, y] : ve) cin >> x >> y;
ll ans = 0;
auto fclamp = [&](ll l, ll r, ll &val) {
if (val < l) { ans += l-val; val = l; }
if (val > r) { ans += val-r; val = r; }
};
for (auto &[x, y] : ve) {
fclamp(1, n, x);
fclamp(1, 2, y);
freq[x][y]++;
}
ll c1 = 0, c2 = 0; // ci = until the current prefix, how many free coins are in row i (negative to show lack of)
for (ll i = 1; i <= n; i++) {
c1 += freq[i][1]-1; // -1 = consumes one
c2 += freq[i][2]-1; // -1 = consumes one
// 4 cases, only in 2 we do something
if (c1 < 0 && c2 > 0) { // shortage in c1 that's satisfied by c2
// ll t = min(-c1, c2);
ll t = 1;
c1 += t; c2 -= t;
ans += t;
} else if (c2 < 0 && c1 > 0) { // vice versa
// ll t = min(-c2, c1);
ll t = 1;
c2 += t; c1 -= t;
ans += t;
}
// ans += abs(val) because if val is positive, #val coins have to move out, if val is negative, #-val coins have to eventually come in
ans += abs(c1);
ans += abs(c2);
}
cout << ans << '\n';
return 0;
}
