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>
using namespace std;
typedef long long ll;
#define sqr(x) ((x)*(x))
#define mp make_pair
#define ld long double
#define PI pair<int,int>
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline ll read(){
	ll x = 0; char ch = gc(); bool positive = 1;
	for (; !isdigit(ch); ch = gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = gc())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(ll a){
    if(a<0){
    	a=-a; putchar('-');
	}
    if(a>=10)write(a/10);
    putchar('0'+a%10);
}
inline void writeln(ll a){write(a); puts("");}
inline void wri(ll a){write(a); putchar(' ');}
const int N=5005;
char s[N],t[N];
int qa[N],qb[N*3];
int main(){
	int n=read();
	scanf("%s%s",s+1,t+1); 
	for(int i=1;i<=n;i++){
		s[i]-='0'; t[i]-='0';
	}
	s[0]=s[n+1]=t[0]=t[n+1]=-1;
	*qa=1;
	for(int i=1;i<=n;i++){
		if(s[i]!=s[i+1])qa[++*qa]=i;
	}
	*qb=*qa+((s[1]!=t[1])^(*qa%2))+1;
	for(int i=1;i<=n;i++){
		if(t[i]!=t[i+1])qb[++*qb]=i;
	}
	for(int i=1;i<=*qa;i++)qb[++*qb]=n;
	int ans=1e9;
	//for(int i=1;i<=*qa;i++)cout<<qa[i]<<" "; puts("");
	//for(int i=1;i<=*qb;i++)cout<<qb[i]<<" "; puts("");
	for(int i=1;i+*qa-1<=*qb;i+=2){
		int sum=0; 
		for(int j=1;j<=*qa;j++)sum+=abs(qa[j]-qb[i+j-1]);
		for(int k=1;k<i;k++)sum+=qb[k];
		for(int k=*qa+i;k<=*qb;k++)sum+=n-qb[k];
		ans=min(ans,sum);
	}
	cout<<ans<<endl;
}