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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,ans=0x3f3f3f3f;
char s[5050],t[5050];
void work(int x)
{
	int res=0;
	for(int i=1;i<x;++i)if(s[i]!=s[i+1])res+=i;
	int p=0;
	for(int i=x;i<=n;++i)
		if(s[i]!=s[i+1])
		{
			if(!p&&t[1]!=s[i+1])res+=i;
			else
			{
				while(t[p]==t[p+1])++p;
				res+=abs(i-p);
				if(p!=n)++p;
			}
		}
	for(int i=p;i<=n;++i)
		if(t[i]!=t[i+1])res+=n-i;
	ans=min(ans,res);
}
int main()
{
	scanf("%d %s %s",&n,s+1,t+1);
	s[0]=s[1]^1,s[n+1]=s[n]^1;
	t[0]=t[1]^1,t[n+1]=t[n]^1;
	for(int i=0;i<=n;++i)if(!i||(s[i-1]!='1'&&s[i]=='1'))work(i);
	swap(s,t);
	for(int i=0;i<=n;++i)if(!i||(s[i-1]!='1'&&s[i]=='1'))work(i);
	printf("%d",ans);
	return 0;
}
