#include <iostream>
#include <vector>
using namespace std;
/*
States:
0 nothing
1 on
2 off
3 toggle
4 on + off
5 on + toggle
6 off + on
7 off + toggle
8 toggle + on
9 toggle + off
We consider the sequence of operations applied to a given position in order.
(*) No two toggle operations can be adjacent.
(*) No operation can be completely contained inside another on/off operation.
*/
const int off = 0;
const int on = 1;
const int toggle = 2;
string s[3] = {"OFF", "ON", "TOGGLE"};
int op_res(int a, int b)
{
if(b == off) return 0;
else if(b == on) return 1;
else return !a;
}
vector<int> ops[10];
int ops_res(int a, int b)
{
for(int x: ops[b]) a = op_res(a, x);
return a;
}
int cost_func(int ops1, int ops2)
{
int ans = ops[ops2].size();
int reducevalue = 0;
if(ops[ops1] == ops[ops2]) reducevalue = ops[ops2].size();
else
{
for(int x: ops[ops1])
for(int y: ops[ops2])
if(x == y)
reducevalue = 1;
}
return ans - reducevalue;
}
const long long INF = 1'000'000'000LL;
long long* dp1;
long long* dp2;
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ops[0] = vector<int>{};
ops[1] = vector<int>{off};
ops[2] = vector<int>{on};
ops[3] = vector<int>{toggle};
ops[4] = vector<int>{off, on};
ops[5] = vector<int>{off, toggle};
ops[6] = vector<int>{on, off};
ops[7] = vector<int>{on, toggle};
ops[8] = vector<int>{toggle, off};
ops[9] = vector<int>{toggle, on};
// for(int i = 0; i <= 9; i++)
// for(int j = 0; j <= 9; j++)
// {
// for(int x: ops[i]) cerr << s[x] << ' ';
// cerr << "-> ";
// for(int y: ops[j]) cerr << s[y] << ' ';
// cerr << ": ";
// cerr << cost(i, j) << '\n';
// }
int N;
cin >> N;
string A;
cin >> A;
A = " " + A;
string B;
cin >> B;
B = " " + B;
dp1 = new long long[10];
dp2 = new long long[10];
dp2[0] = 0;
for(int o = 1; o < 10; o++) dp2[o] = INF;
long long cost[10][10];
for(int o1 = 0; o1 < 10; o1++)
for(int o2 = 0; o2 < 10; o2++)
cost[o1][o2] = cost_func(o1, o2);
for(int i = 1; i <= N; i++)
{
swap(dp1, dp2);
for(int o2 = 0; o2 < 10; o2++)
{
if(ops_res(A[i] - '0', o2) != (B[i] - '0'))
{
dp2[o2] = INF;
}
else
{
dp2[o2] = INF;
for(int o1 = 0; o1 < 10; o1++)
{
dp2[o2] = min(dp2[o2], dp1[o1] + cost[o1][o2]);
}
}
}
}
long long ans = INF;
for(int o = 0; o < 10; o++)
ans = min(ans, dp2[o]);
cout << ans << '\n';
}
