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.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int l = Integer.parseInt(br.readLine());
		int[] a = new int[l];
		for (int i = 0; i < l; i++) {
			a[i] = Integer.parseInt(br.readLine());
		}
		br.close();
		long[] b = new long[l + 1];
		for (int i = 0; i < l; i++) {
			b[i + 1] = b[i] + a[i];
		}

		long ans = b[l];
		long[] dplr = new long[l + 1];
		long[] dpll = new long[l + 1];
		long[] dprr = new long[l + 1];
		for (int i = 0; i < l; i++) {
			dplr[i + 1] = Math.min(Math.min(dplr[i], dprr[i]), b[i]) + (a[i] % 2 ^ 1);
			dpll[i + 1] = Math.min(Math.min(Math.min(dplr[i], dpll[i]), dprr[i]), b[i]) + a[i] % 2;
			dprr[i + 1] = Math.min(dprr[i], b[i]) + a[i] % 2;
			if (a[i] == 0) {
				dpll[i + 1] += 2;
				dprr[i + 1] += 2;
			}
			long val = Math.min(Math.min(dplr[i + 1], dpll[i + 1]), dprr[i + 1]) + b[l] - b[i + 1];
			ans = Math.min(ans, val);
		}
		System.out.println(ans);
	}
}
