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<cstdio>
#include<cstdlib>
#include<cstring>
#define N 200010
using namespace std;
char str1[N],str2[N];
bool num1[N],num2[N];
int t1[N],t2[N],n;
int work()
{
	int tot1=0,tot2=0;
	for(int i=1;i<n;i++) if(!num1[i] && num1[i+1]) t1[++tot1]=i;
	tot2=n;
	memset(t2,0,sizeof(t2));
	for(int i=1;i<n;i++) if(!num2[i] && num2[i+1]) t2[++tot2]=i;
	for(int i=1;i<=n;i++) t2[++tot2]=n;
	int ans=1000000000;
	for(int i=1;i<=tot2-tot1+1;i++)
	{
		int res=0;
		for(int j=1;j<i;j++) res+=t2[j];
		for(int j=1;j<=tot1;j++) res+=abs(t1[j]-t2[i+j-1]);
		for(int j=i+tot1;j<=tot2;j++) res+=n-t2[j];
		ans=min(ans,res);
	}
	return ans;
}
int main()
{
	scanf("%d",&n);
	scanf("%s%s",str1+1,str2+1);
	for(int i=1;i<=n;i++) num1[i]=str1[i]=='1',num2[i]=str2[i]=='1';
	if(n<=2)
	{
		int ans=0;
		for(int i=1;i<=n;i++) ans+=num1[i]!=num2[i];
		printf("%d\n",ans);
		return 0;
	}
	int res=work();
	for(int i=1;i<=n;i++)
	num1[i]=!num1[i],num2[i]=!num2[i];
	res+=work();
	printf("%d\n",res);
	return 0;
}