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<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
int n,_a[10005],_b[10005],*a=_a+5000,*b=_b+5000,n1,n2;
char s[5005],t[5005];
int main()
{
	scanf("%d",&n);
	scanf("%s%s",s+1,t+1);
	if(s[1]!=t[1])a[++n1]=0;
	for(int i=1;i<=n;i++)
	{
		if(s[i]!=s[i+1])a[++n1]=i;
		if(t[i]!=t[i+1])b[++n2]=i;
	}
	int m1=n1,m2=n2;
	while(m1<=5000)a[++m1]=n;
	while(m2<=5000)b[++m2]=n;
	int ans=1000000000;
	for(int i=-n1;i<=n2;i++)
	{
		if(i&1)continue;
		int res=0;
		for(int j=-5000;j<=5000;j++)
		{
			if(j+i>=-5000&&j+i<=5000)res+=abs(a[j+i]-b[j]);
			else
			{
				if(j+i<-5000)res+=b[j];
				else res+=n-b[j];
			}
		}
		ans=min(ans,res);
	}
	printf("%d\n",ans);
	return 0;
}