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;
const int N=5005;
char s1[N],s2[N];
int t1[N],t2[N],tot1,tot2;
int main()
{
	int n,minn=1e9;
	scanf("%d%s%s",&n,s1+1,s2+1);
	t1[++tot1]=0;
	t2[++tot2]=0;
	for(int i=1;i<=n;i++)
	{
		if(s1[i]==s1[i+1]) t1[++tot1]=i;
		if(s2[i]==s2[i+1]) t2[++tot2]=i;
	}
	for(int i=-tot1;i<=tot2;i++)
	{
		if((i&1)==(s1[1]==s2[1])) continue;
		int ans=0;
		for(int j=1;j<=tot1;j++)
		{
			if(i+j>=1&&i+j<=tot2) ans+=abs(t1[j]-t2[i+j]);
			else if(i+j<1) ans+=t1[j];
			else ans+=n-t1[j];
		}
		for(int j=1;j<=tot2;j++)
		{
			if(j-i<1) ans+=t2[j];
			else if(j-i>tot1) ans+=n-t2[j];
		}
		minn=min(minn,ans);
	}
	printf("%d\n",minn);
	return 0;
}