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[]) {
        try (Scanner in = new Scanner(System.in)) {
            int L = in.nextInt();
            int[] A = new int[L];
            for (int i = 0; i < L; i++) {
                A[i] = in.nextInt();
//                A[i] = A[i] % 2 == 0 ? (A[i] == 0 ? 0 : 2) : 1;
            }

            long[][] dp = new long[L + 1][5];
            for (int i = 0; i <= L; i++) {
                if (i == 0) {
                    dp[i][0] = A[i];
                    dp[i][1] = (A[i] == 0 ? 0 : (A[i] % 2 == 0 ? 0 : 1));
                    dp[i][2] = (A[i] == 0 ? 0 : (A[i] % 2 == 0 ? 1 : 0));
                    dp[i][3] = (A[i] == 0 ? 0 : (A[i] % 2 == 0 ? 0 : 1));
                    dp[i][4] = A[i];
//                    Utils.debug("dp[" + i + "]", dp[i]);
                    continue;
                }
                for (int j = 0; j < 5; j++) {
                    if (j == 0) {
                        dp[i][j] = (i < L ? A[i] : 0) + dp[i - 1][j];
                    } else if (j == 1) {
                        dp[i][j] = (i < L ? (A[i] == 0 ? 2 : (A[i] % 2 == 0 ? 0 : 1)) : 0) + Math.min(dp[i - 1][j], dp[i - 1][0]);
                    } else if (j == 2) {
                        dp[i][j] = (i < L ? (A[i] == 0 ? 1 : (A[i] % 2 == 0 ? 1 : 0)) : 0) + Math.min(dp[i - 1][j], Math.min(dp[i - 1][1], dp[i - 1][0]));
                    } else if (j == 3) {
                        dp[i][j] = (i < L ? (A[i] == 0 ? 2 : (A[i] % 2 == 0 ? 0 : 1)) : 0) + Math.min(Math.min(dp[i - 1][3], dp[i - 1][2]), Math.min(dp[i - 1][1], dp[i - 1][0]));
                    } else if (j == 4) {
                        dp[i][j] = (i < L ? A[i] : 0) + Math.min(dp[i - 1][4], Math.min(Math.min(dp[i - 1][3], dp[i - 1][2]), Math.min(dp[i - 1][1], dp[i - 1][0])));
                    }
                }
//                Utils.debug("dp[" + i + "]", dp[i]);
            }

            System.out.println(dp[L][4]);
        }
    }
}
