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.Arrays;
import java.util.Scanner;

public class Main {

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        for(int i=0; i<n; i++) {
            a[i] = sc.nextInt();
        }
        long[][] memo = new long[n+1][5];

        for(int i=0; i<n; i++) {
            int d0 = a[i];
            int d2 = a[i] == 0 ? 2 : (a[i]%2);
            int d1 = (a[i]+1) % 2;
            memo[i+1][0] = memo[i][0] + d0;
            memo[i+1][1] = Math.min(memo[i][0], memo[i][1]) + d2;
            memo[i+1][2] = Math.min(memo[i][0], Math.min(memo[i][1], memo[i][2])) + d1;
            memo[i+1][3] = Math.min(memo[i][0], Math.min(memo[i][1], Math.min(memo[i][2], memo[i][3]))) + d2;
            memo[i+1][4] = Math.min(memo[i][0], Math.min(memo[i][1], Math.min(memo[i][2], Math.min(memo[i][3], memo[i][4])))) + d0;
        }
        Arrays.sort(memo[n]);
        System.out.println(memo[n][0]);

    }

}
