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
//BadWaper gg
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<vector>
#include<deque>
#include<bitset>
#include<map>
#include<set>
#define inf 1e9
#define eps 1e-6
#define mp make_pair
#define N 20010 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
inline ll read()
{
	char ch=getchar();
	ll s=0,w=1;
	while(ch<'0'||ch>'9'){if(ch=='-')w=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){s=s*10+ch-'0';ch=getchar();}
	return s*w;
}
int a[N],b[N];
char s[N],t[N];
int n;
inline int calc()
{
	int t1=0,t2=0;
	for(register int i=1;i<n;i++)if(s[i]=='0'&&s[i+1]=='1')a[++t1]=i;
	for(register int i=1;i<=n;i++)b[++t2]=0;
	for(register int i=1;i<n;i++)if(t[i]=='0'&&t[i+1]=='1')b[++t2]=i;
	for(register int i=1;i<=n;i++)b[++t2]=n;
	int ans=inf;
	for(register int i=1;i<=t2-t1+1;i++)
	{
		int sum=0;
		for(register int j=1;j<i;j++)sum+=b[j];
		for(register int j=1;j<=t1;j++)sum+=abs(a[j]-b[i+j-1]);
		for(register int j=i+t1;j<=t2;j++)sum+=n-b[j];
		ans=min(ans,sum);
	}
	return ans;
}
int main()
{
	//freopen(".in","r",stdin);
	//freopen(".out","w",stdout);
	n=read();
	scanf("%s%s",s+1,t+1);
	if(n<=2)
	{
		int ans=0;
		for(register int i=1;i<=n;i++)ans+=(s[i]!=t[i]);
		printf("%d\n",ans);return 0;
	}
	int ans=calc();
	for(register int i=1;i<=n;i++)s[i]='1'+'0'-s[i];
	for(register int i=1;i<=n;i++)t[i]='1'+'0'-t[i];
	ans=ans+calc();
	printf("%d\n",ans);
	return 0;
}
