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;
typedef long long ll;
const int N=5005,p=1000000007,inv2=(p+1)/2,inf=0x3f3f3f3f;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,cnt1,cnt2,pos1[N],pos2[N],ans;
char a[N],b[N];
int main(){
	n=read();
	scanf("%s %s",a+1,b+1);
	a[0]=b[0]='0';
	for (int i=1;i<=n;i++){
		if (a[i]!=a[i-1]) pos1[++cnt1]=i;
		if (b[i]!=b[i-1]) pos2[++cnt2]=i;
	}
	ans=0x3f3f3f3f;
	for (int d=-cnt2;d<=cnt1;d++)
	if (d%2==0){
		int x=1,y=1;
		if (d<0) y+=d;
		else x-=d;
		int t=0;
		while ((x<=cnt1)||(y<=cnt2)){
			int tx=(x<1) ? 1 : ((x<=cnt1) ? pos1[x] : (n+1));
			int ty=(y<1) ? 1 : ((y<=cnt2) ? pos2[y] : (n+1));
			t+=abs(tx-ty);
			x++;y++;
		}
		ans=min(ans,t);
	}
	printf("%d\n",ans);
	return 0;
}