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
//by yjz
#include<bits/stdc++.h>
using namespace std;
#define FF first
#define SS second
#define PB push_back
#define MP make_pair
#define foreach(it,s) for(__typeof((s).begin()) it=(s).begin();it!=(s).end();it++)
#ifndef LOCAL
#define cerr if(0)cout
#endif
typedef long long ll;
const int mod=1e9+7;
//My i/o stream
struct fastio
{
	char s[100000];
	int it, len;
	fastio() {it=len=0;}
	inline char get()
	{
		if (it<len) return s[it++]; it=0;
		len=fread(s, 1, 100000, stdin);
		if (len==0) return EOF; else return s[it++];
	}
	bool notend()
	{
		char c=get();
		while (c==' '||c=='\n') c=get();
		if (it>0) it--;
		return c!=EOF;
	}
}_buff;
#define geti(x) x=getnum()
#define getii(x,y) geti(x),geti(y)
#define getiii(x,y,z) getii(x,y),geti(z)
#define puti(x) putnum(x),putchar(' ')
#define putii(x,y) puti(x),puti(y)
#define putiii(x,y,z) putii(x,y),puti(z)
#define putsi(x) putnum(x),putchar('\n')
#define putsii(x,y) puti(x),putsi(y)
#define putsiii(x,y,z) putii(x,y),putsi(z)
inline ll getnum()
{
	ll r=0; bool ng=0; char c; c=_buff.get();
	while (c!='-'&&(c<'0'||c>'9')) c=_buff.get();
	if (c=='-') ng=1, c=_buff.get();
	while (c>='0'&&c<='9') r=r*10+c-'0', c=_buff.get();
	return ng?-r:r;
}
template <class T> inline void putnum(T x)
{
	if (x<0) putchar('-'), x=-x;
	register short a[20]={}, sz=0;
	while (x) a[sz++]=x%10, x/=10;
	if(sz==0) putchar('0');
	for (int i=sz-1; i>=0; i--) putchar('0'+a[i]);
}
inline char getreal() {char c=_buff.get(); while (c<=32) c=_buff.get(); return c;}
ll qpow(ll x, ll k) {return k==0? 1: 1ll*qpow(1ll*x*x%mod,k>>1)*(k&1?x:1)%mod;}
const int maxn = 5011;
int n;
int a[maxn], b[maxn];
void input(int a[])
{
	for (int i=0; i<n; i++)
	{
		char c = getreal();
		a[i] = c-'0';
	}
	for (int i=n-1; i>=1; i--) a[i] ^= a[i-1];
}
vector<int> v;
int solve(int k)
{
	int it = 0, ans = 0;
	for (int i=0; i<=n; i++)
	{
		int times;
		if (a[i]==0) times=1;
		if (i==n) times = n;
		if (i==0) times = max(0, k);
		while (times>0&&k<0)
		{
			ans += i;
			k++;
			times--;
		}
		while (times>0&&it<v.size())
		{
			ans += abs(i-v[it++]);
			times--;
		}
		while (i<n&&times>0&&it==v.size())
		{
			ans += abs(i-n);
			times--;
		}
	}
	return ans;
}
int main()
{
	geti(n);
	input(a);
	input(b);
	for (int i=1; i<n; i++) if (b[i]==0) v.PB(i);
	int m = (n+1)/2;
	int ans = 1e9;
	for (int i=-m; i<=m; i++)
	{
		if ((i+a[0]-b[0])%2!=0) continue;
		ans = min(ans, solve(i));
	}
	cout<<ans<<endl;
	return 0;
}
