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<stdio.h>
#include<algorithm>
using namespace std;
const int inf=2147483647;
char s[5010],t[5010];
int n;
int p1[5010],p2[5010],n1,n2;
int res;
void work(char*s,char*t){
	int i,j,u;
	n1=0;
	for(i=0;i<=n;i++){
		if(s[i]!=s[i+1])p1[++n1]=i;
	}
	n2=0;
	if(s[1]!=t[1])p2[++n2]=0;
	for(i=0;i<=n;i++){
		if(t[i]!=t[i+1])p2[++n2]=i;
	}
	if(s[1]==t[n])p2[++n2]=n;
	for(i=1;i<=n2;i+=2){
		u=0;
		for(j=i;j>0;j--)u+=p2[j];
		if(n2-i<n1-1){
			for(j=2;j<=n1;j++)u+=abs(p1[j]-p2[min(n2,i+j-1)]);
		}else{
			for(j=1;j<=n2-i;j++)u+=abs(p1[min(n1,j+1)]-p2[i+j]);
		}
		res=min(res,u);
	}
}
int main(){
	scanf("%d%s%s",&n,s+1,t+1);
	s[0]=s[1]^1;
	s[n+1]=s[n]^1;
	t[0]=t[1]^1;
	t[n+1]=t[n]^1;
	res=inf;
	work(s,t);
	work(t,s);
	printf("%d",res);
}