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>
const int N=5005;
char s1[N],s2[N];
int a[N*2],b[N*2],ans,n,cnt1,cnt2;
int main(){
	//freopen("1_71.txt","r",stdin);  
	scanf("%d",&n);
	scanf("%s%s",s1+1,s2+1);
	for (int i=2;i<=n;i++){
		if (s1[i]!=s1[i-1]) a[++cnt1]=i-1;
		if (s2[i]!=s2[i-1]) b[++cnt2]=i-1;
	}
	if (cnt1>cnt2) std::swap(cnt1,cnt2),std::swap(a,b);
	int i,t=(cnt1/4+1)*2;
	for (int i=1;i<=t;i++) b[++cnt2]=n;
	for (int i=cnt2+t;i>t;i--) b[i]=b[i-t];
	cnt2+=t;
	for (int i=1;i<=t;i++) b[i]=0;
	if (s1[1]==s2[1]) i=1;
	else i=0;
	ans=n*n;
	for (;i+cnt1-1<=cnt2;i+=2){
		int sum=0;
		for (int j=1;j<i;j++) sum=sum+b[j];
		for (int j=1;j<=cnt1;j++) sum=sum+abs(a[j]-b[i+j-1]);
		for (int j=i+cnt1;j<=cnt2;j++) sum=sum+n-b[j];
		ans=std::min(sum,ans);
	}
	printf("%d\n",ans);
}