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 rep(i,n) for ((i)=1;(i)<=(n);(i)++)
using namespace std;
int n,i,j,ans=0x3f3f3f3f;
string s,t;
vector<int> a,b;
int main(){
	cin>>n>>s>>t;
//	s=" "+s;t=" "+t;
	rep(i,n-1){
		if(s[i]!=s[i-1]) a.push_back(i);
	}
	rep(i,n-1){
		if(t[i]!=t[i-1]) b.push_back(i);
	}
//	for(i=0;i<a.size();i++)cerr<<a[i]<<' ';cerr<<endl;
//	for(i=0;i<b.size();i++)cerr<<b[i]<<' ';cerr<<endl;
//	cerr<<a.size()<<' '<<b.size()<<endl;
//	cerr<<-max((int)a.size(),(int)b.size())-1<<' '<<max(a.size(),b.size())+1<<endl;
	for(i=-max((int)a.size(),(int)b.size())-1;;i++){
		if(!(i<=max((int)a.size(),(int)b.size())+1)) break;
		if((i&1)^(s[0]==t[0])){
			int res=0;
			for(j=min(0,-i);j<max((int)a.size(),((int)b.size())-i);j++){
				int k=i+j,px,py;
				px=(j<0?0:(j>=a.size()?n:a[j]));
				py=(k<0?0:(k>=b.size()?n:b[k]));
				res+=abs(px-py);
			}
			ans=min(ans,res);
		}
	}
	cout<<ans<<endl;
	return 0;
}