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>
#include<utility>

using namespace std;

typedef long long ll;

const ll INF = 1145141919810893;

int L;
ll A[200000 + 100];

ll dp[200000 + 100][5];//0　偶　奇　偶　0

int main() {
	cin >> L;
	for (int i = 0; i < L; i++)cin >> A[i];

	for (int i = 0; i <= L; i++)for (int j = 0; j < 5; j++)dp[i][j] = INF;
	dp[0][0] = 0;
	for (int i = 0; i <= L; i++) {
		dp[i + 1][0] = dp[i][0] + A[i];
		dp[i + 1][1] = min({ dp[i][0], dp[i][1] }) + (A[i] % 2 ? 1 :(A[i] != 0 ? 0 : 2));
		dp[i + 1][2] = min({ dp[i][0], dp[i][1], dp[i][2] }) + (A[i] % 2 ? 0 : 1);
		dp[i + 1][3] = min({ dp[i][0], dp[i][1], dp[i][2], dp[i][3] }) + (A[i] % 2 ? 1 : (A[i] != 0 ? 0 : 2));
		dp[i + 1][4] = min({ dp[i][0], dp[i][1], dp[i][2], dp[i][3], dp[i][4] }) + A[i];
	}
	cout << min({ dp[L][0], dp[L][1], dp[L][2], dp[L][3], dp[L][4] }) << endl;
	return 0;
}