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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N=5005;
int n,ans=1<<30;char s[N],t[N];
vector<pair<int,int> >foo,bar;
void solve(){
	for(int i=0;i<foo.size();++i)
		if(foo[i].second==bar[0].second){
			int res=0;
			for(int j=0;j<bar.size();++j)
				res+=abs((i+j<foo.size()?foo[i+j].first:n)-bar[j].first);
			for(int j=0;j<i;++j)
				res+=foo[j].first;
			ans=min(ans,res);
		}
}
int main(){
	scanf("%d%s%s",&n,s+1,t+1);
	if(n<=2){
		ans=0;
		for(int i=1;i<=n;++i)ans+=(s[i]!=t[i]);
		printf("%d\n",ans);
		return 0;
	}
	for(int i=1;i<n;++i){
		if(s[i]!=s[i+1])
			foo.push_back({i,s[i]-'0'});
		if(t[i]!=t[i+1])
			bar.push_back({i,t[i]-'0'});
	}
	if(foo.size()>bar.size())
		swap(foo,bar);
	while(foo.size()<bar.size())
		foo.push_back({n,!foo.back().second});
	solve();swap(foo,bar);solve();
	printf("%d\n",ans);
	return 0;
}
