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.*;
import java.io.*;

import static java.lang.System.in;

public class Main {
    public static void main(String[] args)throws IOException{
        Scanner sc = new Scanner(System.in);
        int L = sc.nextInt();
        long[] A = new long[L+1];
        for(int i=1;i<=L;i++) A[i]=sc.nextLong();
        // leftBack requires come back at i, while left requires not coming back
        if(L==1){
            System.out.println(0);
            return;
        }
        long[] leftSum = new long[L+1];
        for(int i=1;i<=L;i++) leftSum[i] = leftSum[i-1]+A[i];
        long[] leftBack = new long[L+1],left = new long[L+1];
        for(int i=1;i<=L;i++){
            leftBack[i] = Math.min(leftSum[i],leftBack[i-1]+(A[i]==0?2:A[i]%2));
            left[i] = Math.min(leftBack[i-1],left[i-1])+(1-A[i]%2);
        }
        long[] rightSum = new long[L+2];
        for(int i=L;i>=0;i--) rightSum[i] = rightSum[i+1]+A[i];
        long[] rightBack = new long[L+1], right = new long[L+1];
        for(int i=L-1;i>=0;i--){
            rightBack[i] = Math.min(rightSum[i+1],rightBack[i+1]+(A[i+1]==0?2:A[i+1]%2));
            right[i] = Math.min(rightBack[i+1],right[i+1])+(1-A[i+1]%2);
        }
        long ans = Long.MAX_VALUE;
        for(int i=0;i<=L;i++){
            long cur = Math.min(leftBack[i]+right[i],left[i]+rightBack[i]);
            cur = Math.min(leftBack[i]+rightBack[i],cur);
            ans = Math.min(ans,cur);
        }
        System.out.println(ans);
    }
}
