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<vector>
#include<math.h>
using namespace std;
vector<pair<int,int> > pos,pos1;
const int inf=1e9+7;
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n,i,j,k,l,num=0,ans=inf,now;
	cin>>n;
	string s,t;
	cin>>s>>t;
	s=" "+s;
	t=" "+t;
	pos.push_back({0,1-(int)(s[1]-'0')});
	pos1.push_back({0,1-(int)(t[1]-'0')});
	for(i=1;i<n;i++){
		if(s[i]!=s[i+1]){
			pos.push_back({i,(int)(s[i]-'0')});
		}
		if(t[i]!=t[i+1]){
			pos1.push_back({i,(int)(t[i]-'0')});
		}
	}
	pos.push_back({n,(int)(s[n]-'0')});
	pos1.push_back({n,(int)(t[n]-'0')});
//	for(i=0;i<pos.size();i++){
//		cout<<pos[i].first<<' '<<pos[i].second<<endl;
//	}
//	cout<<"cac\n";
//	for(i=0;i<pos1.size();i++){
//		cout<<pos1[i].first<<' '<<pos1[i].second<<endl;
//	}
//	cout<<"cac1\n";
	for(i=0;i<pos1.size();i++){
		if(pos1[i].second!=pos[0].second){
			continue;
		}
		num=0;
		for(now=0;now<i;now++){
			num+=pos1[now].first;
		}
		for(j=0;j<pos.size();j++){
			num+=abs(pos[j].first-pos1[now].first);
			now=min(now+1,(int)pos1.size()-1);
		}
		while(now<pos1.size()-1){
			num+=n-pos1[now].first;
			now++;
		}
		ans=min(ans,num);
	//	cout<<ans<<' '<<i+1<<endl;
	}
	//cout<<ans<<endl;
	for(i=0;i<pos.size();i++){
		if(pos[i].second!=pos1[0].second){
			continue;
		}
		num=0;
		for(now=0;now<i;now++){
			num+=pos[now].first;
		}
		for(j=0;j<pos1.size();j++){
			num+=abs(pos1[j].first-pos[now].first);
			now=min(now+1,(int)pos.size()-1);
		}
		while(now<pos.size()-1){
			num+=n-pos[now].first;
			now++;
		}
		ans=min(ans,num);
	}
	cout<<ans;
}