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=10010;
char s1[N],s2[N];
int _a[N],n1,n2,_b[N],*a=_a+5005,*b=_b+5005,n;
int main(){
	scanf("%d",&n);
	scanf("%s%s",s1+1,s2+1);
	if (s1[1]!=s2[1])a[++n1]=0;
	for (int i=1;i<n;i++)
		if (s1[i]!=s1[i+1])a[++n1]=i;
	for (int i=1;i<n;i++)
		if (s2[i]!=s2[i+1])b[++n2]=i;
	int k1=n1,k2=n2;
	while (k1<=5000)a[++k1]=n;
	while (k2<=5000)b[++k2]=n;
	int ans=1e9;
	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);
}
