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.*;

public class Main {

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

        int l = sc.nextInt();
        int[] a = new int[l];
        for (int i = 0; i < l; i++) {
            a[i] = sc.nextInt();
        }

        // 0 - hidari 0
        // 1 - hidari g
        // 2 - k
        // 3 - migi g
        // 4 - migi 0
        long INF = 2L << 60;
        long[][] dp = new long[l + 1][5];
        for (int i = 1; i < dp.length; i++) {
            for (int j = 0; j < 5; j++) {
                dp[i][j] = INF;
            }
        }

        for (int i = 0; i < l; i++) {

            for (int j = 0; j < 5; j++) {
                long nex = 0;
                if (j == 0 || j == 4) {
                    nex = dp[i][j] + a[i];
                }
                if (j == 1 || j == 3) {
                    if (a[i] % 2 == 0 && a[i] != 0)
                        nex = dp[i][j];
                    else if (a[i] == 0) {
                        nex = dp[i][j] + 2;
                    } else {
                        nex = dp[i][j] + 1;
                    }

                }
                if (j == 2) {
                    if (a[i] % 2 == 1)
                        nex = dp[i][j];
                    else
                        nex = dp[i][j] + 1;
                }

                for (int k = j; k < 5; k++) {
                    dp[i + 1][k] = Math.min(dp[i + 1][k], nex);
                }
            }
        }
        long ans = dp[l][0];
        for (int i = 0; i < 5; i++) {
            ans = Math.min(ans, dp[l][i]);
        }
        System.out.println(ans);
    }

}
