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();
        long[] a = new long[l];
        for(int i=0; i<l; i++) a[i] = sc.nextLong();

        long[][] dp = new long[l+1][5];
        Arrays.fill(dp[0], 0);

        for(int i=0; i<l; i++){
            long min1 = Math.min(dp[i][0], dp[i][1]);
            long min2 = Math.min(dp[i][2], min1);
            long min3 = Math.min(dp[i][3], min2);
            long min4 = Math.min(dp[i][4], min3);
            dp[i+1][0] = dp[i][0] + a[i];
            if(a[i] == 0){
                dp[i+1][1] = min1 + 2;
            }else{
                dp[i+1][1] = min1 + a[i]%2;
            }
            dp[i+1][2] = min2 + (a[i]+1)%2;
            if(a[i] == 0){
                dp[i+1][3] = min3 + 2;
            }else{
                dp[i+1][3] = min3 + a[i]%2;
            }
            dp[i+1][4] = min4 + a[i];
        }

        System.out.println(Arrays.stream(dp[l]).min().getAsLong());
    }
}