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>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
using namespace std;

const int N=5005,inf=1e9;
int p[N],q[N],n,A,B,ans=1e9;
char s[N],t[N];

int solve(int k){
	int res=0;
	rep(i,1,k-1)
		res+=p[i];
	rep(i,1,B){
		if(k<=0)res+=q[i];
		else if(k<=A)res+=abs(p[k]-q[i]);
		else res+=n-q[i];
		k++;
	}
	if(k<=0)k=1;
	while(k<=A)res+=n-p[k++];
	return res;
}

int main(){
	scanf("%d",&n);
	scanf("%s",s+1);
	scanf("%s",t+1);
	rep(i,1,n-1)
		if(s[i]!=s[i+1])p[++A]=i;
	rep(i,1,n-1)
		if(t[i]!=t[i+1])q[++B]=i;
	if(!A&&!B&&s[1]!=t[1])return printf("%d",n),0;
	int o=(s[1]==t[1]?1:0);
	rep(i,-B-1,A+1)if(!((i+inf-o)&1))
		ans=min(ans,solve(i));
	cout<<ans<<endl;
	return 0;
}