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<iostream>
#include<cstdio>
#define Abs(x) (x<0?-(x):(x))
#define FOR(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int N=5050,INF=1000010000;
int n,A,B,ca,cb,ans,a[N],b[N];
char s[N],t[N];
void work(int l,int r){
	if((ca^cb)!=(Abs(1-l)&1)) return;
	int ret=0;
	FOR(i,1,A){
		int x=i-l+1;
		if(x<1) ret+=a[i];
		if(x>B) ret+=n-a[i];
	}
	FOR(i,1,B){
		int x=i+l-1;
		if(x<1) ret+=b[i];
		else if(x>A) ret+=n-b[i];
		else ret+=Abs(b[i]-a[x]);
	}
	ans=min(ans,ret);
	//cout<<l<<' '<<r<<' '<<ans<<'\n';
}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d%s%s",&n,s+1,t+1);
	if(n<=2){
		cout<<(s[1]!=t[1])+(s[2]!=t[2])<<'\n';
		return 0;
	}
	FOR(i,2,n){
		if(s[i-1]!=s[i]) a[++A]=i-1;
		if(t[i-1]!=t[i]) b[++B]=i-1;
	}
	ca=s[a[1]]^48,cb=t[b[1]]^48;
	if(A>B) swap(A,B),swap(a,b);
	ans=INF;
	FOR(i,1-B,A+1) work(i,i+B-1);
	cout<<ans<<'\n';
}