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;

int n,a[15010],b[15010],an,bn,ans,Ans;
char s[5010],t[5010];

int getabs(int x) {return x<0?-x:x;}

int main()
{
	scanf("%d%s%s",&n,s+1,t+1),an=bn=5000,Ans=1000000000;
	for (int i=2; i<=n; i++) if (s[i]==s[i-1]) a[++an]=i-1;
	for (int i=2; i<=n; i++) if (t[i]==t[i-1]) b[++bn]=i-1;
	for (int i=0; i<=5000; i++) a[i]=b[i]=0;
	for (int i=an+1; i<=15000; i++) a[i]=n;
	for (int i=bn+1; i<=15000; i++) b[i]=n;
	for (int i=-n; i<=n; i++)
	{
		ans=0;
		if ((s[1]==t[1])^(i&1)^1) continue;
		for (int j=max(-i,0); j<=15000&&j<=15000-i; j++) ans+=getabs(b[i+j]-a[j]);
		for (int j=max(i,0); j<=15000&&j<=15000+i; j++) ans+=getabs(b[j]-a[j-i]);
		Ans=min(Ans,ans>>1);
	}
	printf("%d\n",Ans);
	return 0;
}