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
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int L = scanner.nextInt();
    int[] a = new int[L + 1];
    for (int i = 1; i <= L; i++) a[i] = scanner.nextInt();

    long[][] dp = new long[L + 1][5];
    for (int i = 1; i <= L; i++) {
      int f = a[i] % 2 == 0 ? 0 : 1;
      if (a[i] == 0) f = 2;
      int g = a[i] % 2 == 1 ? 0 : 1;

      dp[i][0] = dp[i - 1][0] + a[i];
      dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]) + f;
      dp[i][2] = min(dp[i - 1][0], dp[i - 1][1], dp[i - 1][2]) + g;
      dp[i][3] = min(dp[i - 1][0], dp[i - 1][1], dp[i - 1][2], dp[i - 1][3]) + f;
      dp[i][4] = min(dp[i - 1][0], dp[i - 1][1], dp[i - 1][2], dp[i - 1][3], dp[i - 1][4]) + a[i];
    }

    long min = dp[L][0];
    for (int i = 1; i < 5; i++) min = Math.min(min, dp[L][i]);
    System.out.println(min);
  }

  private static long min(long... args) {
    long m = Long.MAX_VALUE;
    for (long a : args) m = Math.min(m, a);
    return m;
  }
}
