/*input
6
011011
110010
17
10110110110110101
01011100100110000
6
011011
110010
// 4
*/
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;
#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"
const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;
template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }
template<typename T>
void print(vector<T> vec, string name = "") {
cout << name;
for (auto u : vec)
cout << u << ' ';
cout << '\n';
}
#pragma GCC optimize("Ofast")
int main() {
ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int N;
str A, B;
cin >> N >> A >> B;
vii dp(N, vi(3, MOD));
// 0 - done
// 1 - ones
// 2 - zeroes
dp[0][0] = (A[0] != B[0]);
dp[0][1] = 1 + (B[0] == '0');
dp[0][2] = 1 + (B[0] == '1');
for (int i = 1; i < N; ++i)
{
// pildom [0]
dp[i][0] = min(dp[i][0], dp[i - 1][0] + (A[i] != B[i] and A[i - 1] == B[i - 1]));
dp[i][0] = min(dp[i][0], dp[i - 1][1] + (A[i] != B[i] and B[i - 1] == '1'));
dp[i][0] = min(dp[i][0], dp[i - 1][2] + (A[i] != B[i] and B[i - 1] == '0'));
// dp[i][0] = min(dp[i][0], dp[i - 1][0] + (A[i] != B[i] and A[i - 1] == B[i - 1]));
// dp[i][0] = min(dp[i][0], dp[i - 1][2] + (B[i] == '1' and B[i - 1] != '1'));
// pildom [1]
dp[i][1] = min(dp[i][1], dp[i - 1][0] + 1 + (B[i] == '0'));
dp[i][1] = min(dp[i][1], dp[i - 1][1] + (B[i] == '0' and B[i - 1] != '0'));
// dp[i][1] = min(dp[i][1], dp[i - 1][2] + (B[i] == '0'));
dp[i][2] = min(dp[i][2], dp[i - 1][0] + 1 + (B[i] == '1'));
dp[i][2] = min(dp[i][2], dp[i - 1][2] + (B[i] == '1' and B[i - 1] != '1'));
}
// print(dp);
// for (int i = 0; i < N; ++i) printf("%d", dp[i][0]); printf("\n");
// for (int i = 0; i < N; ++i) printf("%d", dp[i][1]); printf("\n");
// for (int i = 0; i < N; ++i) printf("%d", dp[i][2]); printf("\n");
printf("%d\n", min(dp[N - 1][0], min(dp[N - 1][1], dp[N - 1][2])));
}
/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/
