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 maxn=20005;
inline int read(){
	char c=getchar();int t=0,f=1;
	while((!isdigit(c))&&(c!=EOF)){if(c=='-')f=-1;c=getchar();}
	while((isdigit(c))&&(c!=EOF)){t=(t<<3)+(t<<1)+(c^48);c=getchar();}
	return t*f;
}
int n;
char s[maxn],t[maxn];
int as[maxn],at[maxn],cs,ct,a[maxn],b[maxn];
signed main(){
	n=read();
	scanf("%s",s+1);
	scanf("%s",t+1);
	if(s[1]=='1')as[++cs]=0;
	for(int i=1;i<n;i++)if(s[i]!=s[i+1])as[++cs]=i;
	if(s[n]=='1')as[++cs]=n;
	if(t[1]=='1')at[++ct]=0;
	for(int i=1;i<n;i++)if(t[i]!=t[i+1])at[++ct]=i;
	if(t[n]=='1')at[++ct]=n;
	for(int i=1;i<=cs;i++)a[i+ct]=as[i];
	for(int i=cs+ct+1;i<=ct+ct+cs;i++)a[i]=n;
	int ans=1e9;
	for(int i=0;i<=ct+cs;i+=2){
		for(int j=1;j<=i;j++)b[j]=0;
		for(int j=1;j<=ct;j++)b[i+j]=at[j];
		for(int j=i+ct+1;j<=cs+ct+ct;j++)b[j]=n;
		int R=0;
		for(int j=1;j<=cs+ct+ct;j++)R+=abs(a[j]-b[j]);
		ans=min(ans,R);
	}
	printf("%d\n",ans);
	return 0;
}
