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 <bits/stdc++.h>
#include <type_traits>
using namespace std;

using ll=int64_t;
#define int ll

#define FOR(i,a,b) for(int i=int(a);i<int(b);i++)
#define REP(i,b) FOR(i,0,b)
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define ALL(x) x.begin(),x.end()
auto& errStream=cerr;
#ifdef LOCAL
#define cerr (cerr<<"-- line "<<__LINE__<<" -- ")
#else
class CerrDummy{}cerrDummy;
template<class T>
CerrDummy& operator<<(CerrDummy&cd,const T&){
	return cd;
}
using charTDummy=char;
using traitsDummy=char_traits<charTDummy>;
CerrDummy& operator<<(CerrDummy&cd,basic_ostream<charTDummy,traitsDummy>&(basic_ostream<charTDummy,traitsDummy>&)){
	return cd;
}
#define cerr cerrDummy
#endif
#define REACH cerr<<"reached"<<endl
#define DMP(x) cerr<<#x<<":"<<x<<endl
#define ZERO(x) memset(x,0,sizeof(x))
#define ONE(x) memset(x,-1,sizeof(x))

using pi=pair<int,int>;
using vi=vector<int>;
using ld=long double;

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
	os<<"("<<p.first<<","<<p.second<<")";
	return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
	os<<"{";
	REP(i,(int)v.size()){
		if(i)os<<",";
		os<<v[i];
	}
	os<<"}";
	return os;
}

ll read(){
	ll i;
	scanf("%"  SCNd64,&i);
	return i;
}

void printSpace(){
	printf(" ");
}

void printEoln(){
	printf("\n");
}

void print(ll x,int suc=1){
	printf("%" PRId64,x);
	if(suc==1)
		printEoln();
	if(suc==2)
		printSpace();
}

string readString(){
	static char buf[3341000];
	scanf("%s",buf);
	return string(buf);
}

char* readCharArray(){
	static char buf[3341000];
	static int bufUsed=0;
	char* ret=buf+bufUsed;
	scanf("%s",ret);
	bufUsed+=strlen(ret)+1;
	return ret;
}

template<class T,class U>
void chmax(T& a,U b){
	if(a<b)
		a=b;
}

template<class T,class U>
void chmin(T& a,U b){
	if(b<a)
		a=b;
}

template<class T>
T Sq(const T& t){
	return t*t;
}

#define CAPITAL
void Yes(bool ex=true){
	#ifdef CAPITAL
	cout<<"YES"<<endl;
	#else
	cout<<"Yes"<<endl;
	#endif
	if(ex)exit(0);
}
void No(bool ex=true){
	#ifdef CAPITAL
	cout<<"NO"<<endl;
	#else
	cout<<"No"<<endl;
	#endif
	if(ex)exit(0);
}

const ll infLL=LLONG_MAX/3;

#ifdef int
const int inf=infLL;
#else
const int inf=INT_MAX/2-100;
#endif

vi Get(){
	string s=readString();
	int n=s.size();
	vi res(n+1,0);
	res[0]=s[0]-'0';
	REP(i,n-1)
		res[i+1]=(s[i]-'0')^(s[i+1]-'0');
	res[n]=s[n-1]-'0';
	
	vi pos;
	REP(i,n+1)
		if(res[i]==0)
			pos.PB(i);
	
	return pos;
}

vi Waf(vi pos){
	int s=pos.size();
	vi res(s+1,inf);
	int cur=0;
	for(int i=0;i<s;i+=2){
		res[i]=cur;
		if(i+1<s)
			cur+=pos[i]+pos[i+1]-1+1;
	}
	if(s%2==0)
		res[s]=cur;
	return res;
}

signed main(){
	int n=read();
	vi s=Get(),t=Get();
	int a=s.size(),b=t.size();
	vi c[2],d[2];
	REP(k,2){
		c[k]=Waf(s);
		d[k]=Waf(t);
		reverse(ALL(s));
		for(auto&v:s)
			v=n-v;
		reverse(ALL(t));
		for(auto&v:t)
			v=n-v;
	}
	int ans=inf;
	{
		int p=inf,q=inf;
		for(int i=0;i<=a;i+=2){
			chmin(p,c[0][i]+c[1][a-i]);
		}
		for(int i=0;i<=b;i+=2){
			chmin(q,d[0][i]+d[1][b-i]);
		}
		chmin(ans,p+q);
	}
	REP(_,2){
		for(int h=0;h<a;h+=2){
			int i=h,j=0;
			int cost=c[0][h];
			while(i<a&&j<b){
				cost+=abs(s[i]-t[j]);
				i++;
				j++;
			}
			cost+=c[1][a-i];
			cost+=d[1][b-j];
			chmin(ans,cost);
		}
		swap(s,t);
		swap(a,b);
		swap(c,d);
	}
	print(ans);
}
