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 <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;

const int N=2e4+5;

int n,a[N],b[N];
char s[N],t[N];

int calc() {
	int t1=0,t2=0;
	fo(i,1,n-1) if (s[i]=='0'&&s[i+1]=='1') a[++t1]=i;
	fo(i,1,n) b[++t2]=0;
	fo(i,1,n-1) if (t[i]=='0'&&t[i+1]=='1') b[++t2]=i;
	fo(i,1,n) b[++t2]=n;
	int ret=n*n;
	fo(i,1,t2-t1+1) {
		int now=0;
		fo(j,1,i-1) now+=b[j];
		fo(j,1,t1) now+=abs(a[j]-b[i+j-1]);
		fo(j,i+t1,t2) now+=n-b[j];
		ret=min(ret,now);
	}
	return ret;
}

int main() {
	scanf("%d",&n);scanf("%s",s+1);scanf("%s",t+1);
	if (n<=2) {
		int ans=0;
		fo(i,1,n) ans+=s[i]!=t[i];
		printf("%d\n",ans);
		return 0;
	}
	int tmp=calc();
	fo(i,1,n) s[i]='0'+'1'-s[i],t[i]='0'+'1'-t[i];
	printf("%d\n",tmp+calc());
	return 0;
}