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<algorithm>
#define RI register int
#define CI const int&
using namespace std;
const int N=5005;
int n,sa[N<<2],sb[N<<2],ans; char s[N],t[N];
inline void calc(int ret=1e9)
{
	RI i,j,ta=0,tb=0; for (i=1;i<=n;++i) sa[++ta]=0;
	for (i=1;i<n;++i) if (s[i]=='1'&&s[i+1]=='0') sa[++ta]=i;
	for (i=1;i<=n;++i) sa[++ta]=n;
	for (i=1;i<n;++i) if (t[i]=='1'&&t[i+1]=='0') sb[++tb]=i;
	for (i=1;i+n-1<=ta;++i)
	{
		int cur=0; for (j=1;j<i;++j) cur+=sa[j];
		for (j=1;j<=tb;++j) cur+=abs(sa[i+j-1]-sb[j]);
		for (j=i+tb;j<=ta;++j) cur+=n-sa[j]; ret=min(ret,cur);
	}
	ans+=ret;
}
int main()
{
	RI i; scanf("%d%s%s",&n,s+1,t+1);
	if (n==1) return puts(s[1]==t[1]?"0":"1"),0;
	if (n==2&&s[1]!=t[1]&&s[1]==s[2]&&t[1]==t[2]) return puts("2"),0;
	for (calc(),i=1;i<=n;++i) s[i]=48+((s[i]-48)^1),t[i]=48+((t[i]-48)^1);
	return calc(),printf("%d",ans),0;
}