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>

const int N = 5000 + 7;

int n;
int a[N], b[N];
int xa[3], xb[3];
int ans;

int abs(int x) {
	return x < 0 ? -x : x;
}
void calc(int x, int y) {
	if (x < 0 || y < 0) return;
	int cur = 0;
	for (int i = x, j = y; i < n || j < n; ) {
		if (i < n && a[i] == -1) { ++i; continue; }
		if (j < n && b[j] == -1) { ++j; continue; }
		cur += abs(i - j), i += i < n, j += j < n;
	}
	for (int i = x - 1, j = y - 1; i > 0 || j > 0; ) {
		if (i > 0 && a[i] == -1) { --i; continue; }
		if (j > 0 && b[j] == -1) { --j; continue; }
		cur += abs(i - j), i -= i > 0, j -= j > 0;
	}
	ans = std::min(ans, cur);
}

int main() {
//	freopen("code.in", "r", stdin);
//	freopen("code.out", "w", stdout);

	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) scanf("%1d", a + i);
	for (int i = 1; i <= n; ++i) scanf("%1d", b + i);
	for (int i = 1; i <= n; ++i)
		ans += abs(a[i] - b[i]);
	if (ans && n > 2) ans = 1 << 29;
	for (int i = 1; i < n; ++i)
		a[i] = a[i] == a[i + 1] ? -1 : a[i],
		b[i] = b[i] == b[i + 1] ? -1 : b[i];
	xa[0] = xa[1] = xb[0] = xb[1] = -1;
	for (int i = n; i; --i) {
		if (a[i] != -1) xa[a[i]] = i;
		if (b[i] != -1) xb[b[i]] = i;
	}
	for (int i = 1; i < n; ++i) {
		if (a[i] != -1) calc(i, xb[a[i]]);
		if (b[i] != -1) calc(xa[b[i]], i);
	}
	printf("%d", ans);
	return 0;
}
