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>
using namespace std;
const int N=10005,inf=1e9+7;
int n,s1,s2,ans;
char s[N],t[N];
vector<int>a,b;
void sol(vector<int>&V,char *s){for(int i=1;i<n;i++)if(s[i]!=s[i+1])V.push_back(i);}
int main()
{
	scanf("%d%s%s",&n,s+1,t+1);
	sol(a,s);s1=a.size();
	sol(b,t);s2=b.size();
	ans=inf;
	for(int d=-s1-1;d<=s2+1;d++)if((d+(s[1]==t[1]))&1)
	{
		int r=0;
		for(int i=min(-d,0);i<max(s1,s2-d);i++)
		{
			int j=i+d,px,py;
			px=i<0?0:(i>=s1?n:a[i]);
			py=j<0?0:(j>=s2?n:b[j]);
			r+=abs(px-py);
		}
		ans=min(ans,r);
	}
	printf("%d\n",ans);
	return 0;
}