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;
template<typename T>inline T read(){
	T f=0,x=0;char c=getchar();
	while(!isdigit(c)) f=c=='-',c=getchar();
	while(isdigit(c)) x=x*10+c-48,c=getchar();
	return f?-x:x;
}
namespace run{
	inline char getc(){
		char c=getchar();
		while(c!='0' && c!='1') c=getchar();
		return c;
	}

	const int N=5009;
	int s[N],t[N],n,a[N],b[N],top1,top2;
	int main(){
		n=read<int>();
		for(int i=1;i<=n;i++) s[i]=getc()-'0';
		for(int i=1;i<=n;i++) t[i]=getc()-'0';
		for(int i=1;i<n;i++) if(s[i]^s[i+1]) a[++top1]=i;
		for(int i=1;i<n;i++) if(t[i]^t[i+1]) b[++top2]=i;

		int ans=n*n;
		for(int i=-top1-1;i<=top2+1;i++) if(s[1]^t[1]^(~i&1)){
			int ret=0;
			for(int j=min(1,1-i);j<=max(top1,top2-i);j++)
				ret+=abs(((j<0)?0:((j>top1)?n:a[j]))-((j+i<0)?0:((j+i>top2)?n:b[j+i])));
			ans=min(ans,ret);
		}
		printf("%d\n",ans);
		return 0;
	}
}
int main(){
#ifdef my
	freopen(".in","r",stdin);
	freopen(".out","w",stdout);
#endif
	return run::main();
}