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;
const int N=5100;
const int inf=0x3f3f3f3f;
char s[N],t[N];
int n,ans,a[N],b[N],sum1[N],sum2[N];
int main()
{
	scanf("%d%s%s",&n,s+1,t+1);
	if (s[1]!=t[1]) a[++*a]=0; 
	for (int i=1;i<n;i++)
	  if (s[i]!=s[i+1]) a[++*a]=i,sum1[*a]=sum1[*a-1]+i;
	for (int i=1;i<n;i++)
	  if (t[i]!=t[i+1]) b[++*b]=i,sum2[*b]=sum2[*b-1]+i;
	ans=inf;
	for (int i=0;i<=*a;i+=2)
	{
		int x=i+1,y=1,tmp=sum1[i];
		while (x<=*a||y<=*b){
			if (x<=*a&&y<=*b) tmp+=abs(a[x]-b[y]),x++,y++;
			else if (x<=*a) tmp+=n-a[x],x++;
			else if (y<=*b) tmp+=n-b[y],y++;
		}
		ans=min(ans,tmp);
	}
	for (int i=0;i<=*b;i+=2)
	{
		int x=1,y=i+1,tmp=sum2[i];
		while (x<=*a||y<=*b){
			if (x<=*a&&y<=*b) tmp+=abs(a[x]-b[y]),x++,y++;
			else if (x<=*a) tmp+=n-a[x],x++;
			else if (y<=*b) tmp+=n-b[y],y++;
		}
		ans=min(ans,tmp);
	}
	printf("%d\n",ans);
	return 0;
}