You are given strings s and t, both of length N. s and t consist of `0` and `1`. Additionally, in these strings, the same character never occurs three or more times in a row.

You can modify s by repeatedly performing the following operation:

* Choose an index i (1 \leq i \leq N) freely and invert the i-th character in s (that is, replace `0` with `1`, and `1` with `0`), under the condition that the same character would not occur three or more times in a row in s after the operation.



Your objective is to make s equal to t. Find the minimum number of operations required.

Constraints

* 1 \leq N \leq 5000
* The lengths of s and t are both N.
* s and t consists of `0` and `1`.
* In s and t, the same character never occurs three or more times in a row.

Input

Input is given from Standard Input in the following format:


N
s
t


Output

Find the minimum number of operations required to make s equal to t. It can be proved that the objective is always achievable in a finite number of operations.

Examples

Input

4
0011
0101


Output

4


Input

1
0
0


Output

0


Input

8
00110011
10101010


Output

10
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <time.h>
#include <stdlib.h>
#include <string>
#include <bitset>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <algorithm>
#include <sstream>
#include <stack>
#include <iomanip>
using namespace std;
#define pb push_back
#define mp make_pair
typedef pair<int,int> pii;
typedef long long ll;
typedef double ld;
typedef vector<int> vi;
#define fi first
#define se second
#define fe first
#define FO(x) {freopen(#x".in","r",stdin);freopen(#x".out","w",stdout);}
#define Edg int M=0,fst[SZ],vb[SZ],nxt[SZ];void ad_de(int a,int b){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;}void adde(int a,int b){ad_de(a,b);ad_de(b,a);}
#define Edgc int M=0,fst[SZ],vb[SZ],nxt[SZ],vc[SZ];void ad_de(int a,int b,int c){++M;nxt[M]=fst[a];fst[a]=M;vb[M]=b;vc[M]=c;}void adde(int a,int b,int c){ad_de(a,b,c);ad_de(b,a,c);}
#define es(x,e) (int e=fst[x];e;e=nxt[e])
#define esb(x,e,b) (int e=fst[x],b=vb[e];e;e=nxt[e],b=vb[e])
#define SZ 66666
int n,ans=2e9;
char s[SZ],t[SZ];
void wo(vector<pii> a,vector<pii> b)
{
	if(a.size()>b.size())
	{
		wo(b,a);
		return;
	}
	reverse(a.begin(),a.end());
	while(a.size()<b.size()) a.pb(mp(0,!a.back().se));
	reverse(a.begin(),a.end());
	if(a.back().se!=b.back().se) return;
	if(a.size()!=b.size()) throw "GG";
	int su=0;
	for(int i=0;i<a.size();++i)
		su+=abs(a[i].fi-b[i].fi);
	ans=min(ans,su);
}
int main()
{
	cin>>n>>(s+1)>>(t+1);
	for(int i=1;i<=n;++i) s[i]-='0',t[i]-='0';
	if(n<=2)
	{
		int su=0;
		for(int i=1;i<=n;++i)
			su+=s[i]!=t[i];
		printf("%d\n",su);
		return 0;
	}
	vector<pii> p,q,qq;
	for(int i=1;i<n;++i)
	{
		if(s[i]!=s[i+1]) p.pb(mp(i,s[i]));
		if(t[i]!=t[i+1]) q.pb(mp(i,t[i]));
	}
	qq=q;
	for(int i=0;i<=n+10;++i)
	{
		wo(p,qq); qq.pb(mp(n,!qq.back().se));
	}
	for(int i=0;i<=n+10;++i)
	{
		wo(p,q); p.pb(mp(n,!p.back().se));
	}
	cout<<ans<<"\n";
}
