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 _ = 1e5 + 7;
char A[_] , B[_]; deque < int > posA , posB; int N;

int main(){
	scanf("%d %s %s" , &N , A + 1 , B + 1);
	if(N == 1){printf("%d\n" , A[1] ^ B[1]); return 0;}
	for(int i = 2 ; i <= N ; ++i) if(!(A[i] ^ A[i - 1])) posA.push_back(i);
	for(int i = 2 ; i <= N ; ++i) if(!(B[i] ^ B[i - 1])) posB.push_back(i);
	long long ans = 1e18;
	for(int i = -(int)max(posA.size() , posB.size()) - 10 ; i <= (int)max(posA.size() , posB.size()) + 10 ; ++i)
		if(!((A[1] - B[1] + i) & 1)){
			deque < int > pA = posA , pB = posB; long long sum = 0;
			for(int j = 0 ; j < abs(i) ; ++j) (i < 0 ? pA : pB).push_front(1);
			while(pA.size() < pB.size()) pA.push_back(N + 1);
			while(pB.size() < pA.size()) pB.push_back(N + 1);
			for(int j = 0 ; j < pA.size() ; ++j) sum += abs(pA[j] - pB[j]);
			ans = min(ans , sum);
		}
	cout << ans; return 0;
}