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 title "title"
#define ll long long
#define ull unsigned ll
#define fix(x) fixed<<setprecision(x)
#define pii pair<int,int>
#define vint vector<int>
#define pb push_back
using namespace std;
void Freopen(){
	freopen(title".in","r",stdin);
	freopen(title".out","w",stdout);
}
int read(){
	int g=0,f=1;
	char ch=getchar();
	while(ch<'0'||'9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch&&ch<='9'){g=g*10+ch-'0';ch=getchar();}
	return g*f;
}
const int N=5005;
const int INF=1e9;
int n,na,nb,ans=INF;
pii a[N],b[N];
char s[N],t[N];
void solve(){
	for(int i=0;i<=n;i++){
		if((a[1].second+b[1].second+i)&1)continue;
		int re=0,now=1;
		while(now<=na){
			int pos=a[now].first,pre=now+i<=nb?b[now+i].first:n+1;
			re+=abs(pos-pre);now++;
		}
		while(now+i<=nb)re+=n+1-b[now+i].first,now++;
		for(int j=1;j<=i;j++)re+=(j<=nb?b[j].first:n+1)-1;
		ans=min(ans,re);
	}
}
signed main(){
	n=read();scanf("%s%s",s+1,t+1);
	for(int i=1;i<=n;i++){
		if(s[i]!=s[i-1])a[++na]=pii(i,s[i]-'0');
		if(t[i]!=t[i-1])b[++nb]=pii(i,t[i]-'0');
	}solve(),swap(na,nb),swap(a,b),solve();
	return cout<<ans,signed();
}