#include <bits/stdc++.h>
using namespace std;
const int DIM = 100005;
int cnt[DIM][3];
vector<int> qry[3], bal[3];
int main(void) {
#ifdef HOME
freopen("coins.in", "r", stdin);
freopen("coins.out", "w", stdout);
#endif
int n; cin >> n;
long long ans = 0;
for (int i = 1; i <= n * 2; ++i) {
int x, y; cin >> x >> y;
if (x < 1) { ans += 1 - x; x = 1; }
if (x > n) { ans += x - n; x = n; }
if (y < 1) { ans += 1 - y; y = 1; }
if (y > 2) { ans += y - 2; y = 2; }
++cnt[x][y]; }
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= 2; ++j) {
qry[j].push_back(i);
for (; cnt[i][j]; --cnt[i][j]) {
bal[j].push_back(i); } }
// solve as many columns with balls placed on the same line
for (int j = 1; j <= 2; ++j) {
while (qry[j].size() and bal[j].size()) {
int p1 = qry[j].back(), p2 = bal[j].back();
qry[j].pop_back(); bal[j].pop_back(); ans += abs(p1 - p2); } }
// solve as many columns with balls placed on the opposite line
for (int j = 1; j <= 2; ++j) {
while (qry[j].size() and bal[3 - j].size()) {
int p1 = qry[j].back(), p2 = bal[3 - j].back();
qry[j].pop_back(); bal[3 - j].pop_back(); ans += abs(p1 - p2) + 1; } } }
cout << ans << endl;
return 0; }
