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>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<map>
#include<bitset>
#include<set>
#define N (100010)
#define P ()
#define M ()
#define inf (0x3fffffff)
#define rg register int
#define Label puts("NAIVE")
#define spa print(' ')
#define ent print('\n')
#define rand() (((rand())<<(15))^(rand()))
#define file(s) freopen(s".in","r",stdin),freopen(s".out","w",stdout)
typedef long double ld;
typedef long long LL;
typedef unsigned long long ull;
using namespace std;
namespace fastIO1{
	inline char read(){
		static const int IN_LEN=1000000;
		static char buf[IN_LEN],*s,*t;
		return (s==t?t=(s=buf)+fread(buf,1,IN_LEN,stdin),(s==t?-1:*s++):*s++);
	}
	template<class T>
	inline void read(T &x){
		static bool iosig;
		static char c;
		for(iosig=false,c=read();!isdigit(c);c=read()){
			if(c=='-')iosig=true;
			if(c==-1)return;
		}
		for(x=0;isdigit(c);c=read())x=((x+(x<<2))<<1)+(c^'0');
		if(iosig)x=-x;
	}
	inline char readc(char &c){
		for(c=read();!isalpha(c)&&!isdigit(c);c=read())
		if(c==-1)return 0;
	}
	const int OUT_LEN = 10000000;
	char obuf[OUT_LEN],*ooh=obuf;
	inline void print(char c){
		if(ooh==obuf+OUT_LEN)fwrite(obuf,1,OUT_LEN,stdout),ooh=obuf;
		*ooh++=c;
	}
	template<class T>
	inline void print(T x){
		static int buf[30],cnt;
		if(x==0)print('0');
		else{
			if(x<0)print('-'),x=-x;
			for(cnt=0;x;x/=10)buf[++cnt]=x%10+48;
			while(cnt)print((char)buf[cnt--]);
		}
	}
	inline void flush(){fwrite(obuf,1,ooh-obuf,stdout);}
}
namespace fastIO2{
	template<class T>
	inline void read(T &x){
		static bool iosig;
		static char c;
		for(iosig=false,c=getchar();!isdigit(c);c=getchar()){
			if(c=='-')iosig=true;
			if(c==-1)return;
		}
		for(x=0;isdigit(c);c=getchar())x=((x+(x<<2))<<1)+(c^'0');
		if(iosig)x=-x;
	}
}
using namespace fastIO1;
int n,m,a[N],b[N],ans; char s[N],t[N];
void solve(){
	int cnt1=0,cnt2=n,mn=inf;
	memset(b,0,sizeof(b));
	for(int i=1;i<n;i++)if(s[i]!=s[i+1]&&s[i]=='0')a[++cnt1]=i;
	for(int i=1;i<n;i++)if(t[i]!=t[i+1]&&t[i]=='0')b[++cnt2]=i;
	for(int i=1;i<=n;i++)b[++cnt2]=n;
	for(int i=1;i<=cnt2-cnt1+1;i++){
		int S=0; for(int j=1;j<i;j++)S+=b[j];
		for(int j=i;j<=i+cnt1-1;j++)S+=abs(b[j]-a[j-i+1]);
		for(int j=i+cnt1;j<=cnt2;j++)S+=(n-b[j]);
		mn=min(mn,S); 
	}
	ans+=mn;
}
void rev(){for(int i=1;i<=n;i++)s[i]^=1,t[i]^=1;}
int main(){
	read(n);
	for(int i=1;i<=n;i++)readc(s[i]);
	for(int i=1;i<=n;i++)readc(t[i]);
	if(n<=2){for(int i=1;i<=n;i++)ans+=(int)(s[i]!=t[i]); cout<<ans,exit(0);}
	solve(),rev(),solve(),cout<<ans;
}