#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF = 1e9+7;
const ll MOD = 998244353;
typedef pair<ll,ll> ii;
#define iii pair<ii,ll>
#define f(i,a,b) for(ll i = a;i < b;i++)
#define pb push_back
#define vll vector<ll>
#define F first
#define S second
#define all(x) (x).begin(), (x).end()
///I hope I will get uprating and don't make mistakes
///I will never stop programming
///sqrt(-1) Love C++
///Please don't hack me
///@TheofanisOrfanou Theo830
///Think different approaches (bs,dp,greedy,graphs,shortest paths,mst)
///Stay Calm
///Look for special cases
///Beware of overflow and array bounds
///Think the problem backwards
///Training
ll dp[1000005][3][2];
string a,b;
ll n;
ll solve(ll idx,ll j1,ll j2){
if(dp[idx][j1][j2] != -1){
return dp[idx][j1][j2];
}
if(idx == n){
return 0;
}
ll ans = INF;
f(k1,0,3){
f(k2,0,2){
char v = a[idx];
if(k1){
v = k1 - 1 + '0';
}
if(k2){
v = ((v - '0') ^ 1) + '0';
}
if(b[idx] == v){
ans = min(ans,solve(idx+1,k1,k2) + (k1 != j1 && k1 != 0) + (k2 != j2 && k2 != 0));
}
}
}
return dp[idx][j1][j2] = ans;
}
int main(void){
ios_base::sync_with_stdio(0);
cin.tie(0);
cin>>n;
cin>>a>>b;
memset(dp,-1,sizeof dp);
cout<<solve(0,0,0)<<"\n";
}
