Snuke stands on a number line. He has L ears, and he will walk along the line continuously under the following conditions:

* He never visits a point with coordinate less than 0, or a point with coordinate greater than L.
* He starts walking at a point with integer coordinate, and also finishes walking at a point with integer coordinate.
* He only changes direction at a point with integer coordinate.



Each time when Snuke passes a point with coordinate i-0.5, where i is an integer, he put a stone in his i-th ear.

After Snuke finishes walking, Ringo will repeat the following operations in some order so that, for each i, Snuke's i-th ear contains A_i stones:

* Put a stone in one of Snuke's ears.
* Remove a stone from one of Snuke's ears.



Find the minimum number of operations required when Ringo can freely decide how Snuke walks.

Constraints

* 1 \leq L \leq 2\times 10^5
* 0 \leq A_i \leq 10^9(1\leq i\leq L)
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


L
A_1
:
A_L


Output

Print the minimum number of operations required when Ringo can freely decide how Snuke walks.

Examples

Input

4
1
0
2
3


Output

1


Input

8
2
0
0
2
1
3
4
1


Output

3


Input

7
314159265
358979323
846264338
327950288
419716939
937510582
0


Output

1
#include <iostream>
#include <algorithm>
using namespace std;

long long L, A[1 << 18], dp[1 << 18][5], S[1 << 18][5];

int main() {
	cin >> L;
	for (int i = 1; i <= L; i++) {
		cin >> A[i];
		S[i][0] = A[i];
		S[i][4] = A[i];
		S[i][2] = ((A[i] % 2) ^ 1);
		if (A[i] == 0) {
			S[i][1] = 2;
			S[i][3] = 2;
		}
		else {
			S[i][1] = (A[i] % 2);
			S[i][3] = (A[i] % 2);
		}
	}

	for (int i = 0; i <= L; i++) {
		for (int j = 0; j <= 4; j++) dp[i][j] = (1LL << 60);
	}
	dp[0][0] = 0;
	for (int i = 1; i <= L; i++) {
		for (int j = 1; j <= 4; j++) dp[i - 1][j] = min(dp[i - 1][j], dp[i - 1][j - 1]);
		for (int j = 0; j <= 4; j++) {
			dp[i][j] = min(dp[i][j], dp[i - 1][j] + S[i][j]);
		}
	}

	long long ans = (1LL << 60);
	for (int i = 0; i <= 4; i++) ans = min(ans, dp[L][i]);
	cout << ans << endl;
	return 0;
}