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
//problem:agc030_e
#include <bits/stdc++.h>
using namespace std;

#define pf push_front
#define pb push_back
#define mk make_pair
#define lob lower_bound
#define upb upper_bound
#define fst first
#define scd second

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

inline int read(){
	int f=1,x=0;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline ll readll(){
	ll f=1,x=0;char ch=getchar();
	while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
	while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}

const int MAXN=5005,INF=1e9;
int n;
char s[MAXN],t[MAXN];
deque<int>a,b;
int main() {
	scanf("%d%s%s",&n,s+1,t+1);
	for(int i=1;i<=n;++i)s[i]-='0',t[i]-='0';
	s[0]=s[1]^1,s[n+1]=s[n]^1,t[0]=t[1]^1,t[n+1]=t[n]^1;
	for(int i=1;i<=n+1;++i){
		if(s[i]!=s[i-1])a.pb(i);
		if(t[i]!=t[i-1])b.pb(i);
	}
	if(s[1]!=t[1])a.pf(1);
	for(int i=0;i<=n+1;++i)a.pf(1),a.pf(1);
	for(int i=0;i<=n+1;++i)a.pb(n+1),a.pb(n+1);
	int ans=INF;
	for(int i=0;i+b.size()<=a.size();i+=2){
		int tmp=0;
		for(int j=0;j<i;++j)tmp+=abs(a[j]-1);
		for(int j=i;j<(int)(i+b.size());++j)tmp+=abs(a[j]-b[j-i]);
		for(int j=i+b.size();j<(int)a.size();++j)tmp+=abs(a[j]-(n+1));
		ans=min(ans,tmp);
	}
	cout<<ans<<endl;
	return 0;
}