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;

#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define lg long long
#define db double
#define lb(x) ((x)&-(x))
#define ft first
#define sd second

#define HII cerr<<"HI"<<endl
#define LLLINE cerr<<"@@@@@@@@@@@@@@@@@@@@"<<endl

template <class _T_>
void read(_T_& d){
	d=0;int f=1;char c=getchar();
	for(;c<'0'||c>'9';c=getchar())if(c=='-')f*=-1;
	for(;c>='0'&&c<='9';c=getchar())d=d*10+c-'0';
	d*=f;
}

/************************************************/
#define MN 20005
int n;
char s[MN],t[MN];
int pos[MN];int st[MN];
int res=0x3f3f3f3f;
int G1=0,G2=0;
void ck(int x){
	//cerr<<x<<endl;
	//cerr<<pos[x]<<endl;
	int R=0;
	for(int j=-n;j<=n+n;++j){
		int X=min(G1,max(0,x+j));
		int Y=min(G2,max(0,j));
		//cerr<<X<<' '<<pos[X]<<"@"<<Y<<' '<<st[Y]<<endl;
		R+=abs(pos[X]-st[Y]);
	}
	//cerr<<R<<"@"<<endl;
	res=min(R,res);
}
int main(){
	read(n);
	scanf("%s",s+1);
	scanf("%s",t+1);
	for(int i=2;i<=n;++i)if(s[i]!=s[i-1])pos[++G1]=i-1;pos[++G1]=n;
	for(int i=2;i<=n;++i)if(t[i]!=t[i-1])st[++G2]=i-1;st[++G2]=n;
	int H=max(G1,G2);
	for(int i=-H;i<=H;++i)if(s[1]==t[1]^(i&1))ck(i);
	cout<<res;
	return 0;
}
