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 {
	Scanner sc = new Scanner(System.in) ;
	public static void main(String[]args) {
		Main m = new Main();
		m.run();
	}
	
	void run() {
		int l = sc.nextInt();
		
		long[] a = new long[l];
		for(int i = 0 ; i < l ; i++) {
			a[i] = sc.nextInt();
		}
		
		
		long[][] dp = new long[l+1][5];
		
		for(int i=0 ; i<l ; i++) {
			
			dp[i+1][0] = dp[i][0] + a[i];

			long p2 = a[i] == 0 ? 2 : a[i] % 2;
			dp[i+1][1] = min(dp[i], 2) + p2;
			dp[i+1][2] = min(dp[i], 3) + (a[i] + 1)% 2;
			dp[i+1][3] = min(dp[i], 4) + p2;
			dp[i+1][4] = min(dp[i], 5) + a[i];
		}
		
		System.out.println(min(dp[l],5));
		
	}
	
	long min(long[] ls, int to) {
		long res = Long.MAX_VALUE/2;
		for(int i = 0 ; i < to ; i ++) {
			res = Math.min(res, ls[i]);
		}
		return res;
	}
}
