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) {
		Scanner sc = new Scanner(System.in);
		int l = sc.nextInt();
		long dp1 = 0;
		long dp2 = 0;
		long dp3 = 0;
		long dp4 = 0;
		long dp5 = 0;
		long dp6 = 0;
		long dp7 = 0;
		long dp8 = 0;
		long dp9 = 0;
		long dp10 = 0;
		
		for (int i=0;i<l;i++) {
			int a = sc.nextInt();
			dp6 = dp1+a;
			dp7 = Math.min(dp1, dp2)+(a==0? 2: a%2);
			dp8 = Math.min(dp1, Math.min(dp2, dp3))+(a+1)%2;
			dp9 = Math.min(Math.min(dp1, dp2), Math.min(dp3, dp4))+(a==0? 2: a%2);
			dp10 = Math.min(Math.min(dp1, Math.min(dp2, dp3)), Math.min(dp4, dp5))+a;
			
			dp1 = dp6;
			dp2 = dp7;
			dp3 = dp8;
			dp4 = dp9;
			dp5 = dp10;
		}
		System.out.println(Math.min(Math.min(dp1, Math.min(dp2, dp3)), Math.min(dp4, dp5)));
	}


}



