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 MAXN=20100;
char s[MAXN],t[MAXN];
int n,a[MAXN],b[MAXN],sum;
inline int calc()
{
	int cnt1=0,cnt2=0,ans=n*n;
	for(int i=1;i<n;i++)
		if(s[i]=='0'&&s[i+1]=='1') a[++cnt1]=i;
	for(int i=1;i<=n;i++) b[++cnt2]=0;
	for(int i=1;i<n;i++)
		if(t[i]=='0'&&t[i+1]=='1') b[++cnt2]=i;
	for(int i=1;i<=n;i++) b[++cnt2]=n;
	for(int i=1;i<=cnt2-cnt1+1;i++)
	{
		int tmp=0;
		for(int j=1;j<i;j++) tmp+=b[j];
		for(int j=1;j<=cnt1;j++) tmp+=abs(a[j]-b[i+j-1]);
		for(int j=cnt1+i;j<=cnt2;j++) tmp+=n-b[j];
		ans=min(ans,tmp);
	}
	return ans;
}
int main()
{
	ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
	cin>>n>>(s+1)>>(t+1);
	if(n<=2)
	{
		for(int i=1;i<=n;i++) sum+=(s[i]!=t[i]);
		cout<<sum<<endl;
		return 0;
	}
	sum=calc();
	for(int i=1;i<=n;i++) s[i]='0'+'1'-s[i],t[i]='0'+'1'-t[i];
	sum+=calc();
	cout<<sum<<endl;
	return 0;
}