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.Arrays;
import java.util.Scanner;

public class Main {
	
	public static void main(String[] args)
	{
		new Main().run();
	}
	
	public void run()
	{
		Scanner file = new Scanner(System.in);
		int L = file.nextInt();
		int[] ints = new int[L+2];//sentinal first and last
		int[][] pred = new int[][]{{0},{0,1},{0,1,2},{2,3},{0,1,2,3,4}};
		for(int i = 0;i<L;i++)
			ints[i+1] = file.nextInt();
		long[][] dp = new long[L+2][5];
		for(long[] in:dp)
			Arrays.fill(in, Long.MAX_VALUE/2);
		dp[0][0] = 0;
		for(int i = 1;i<dp.length;i++)
		{
			for(int j = 0;j<dp[0].length;j++)
			{
				long minPred = Long.MAX_VALUE/2;
				for(int k:pred[j])
					minPred = Math.min(minPred,dp[i-1][k]);
				if(j==0 || j == 4)
				{
					dp[i][j] = Math.min(dp[i][j], minPred + ints[i]);
				}else if(j == 1 || j == 3)
				{//even start or even end
					if(ints[i] == 0)
						dp[i][j] = Math.min(dp[i][j], minPred + 2);
					else
						dp[i][j] = Math.min(dp[i][j], minPred + ints[i] % 2);
				}else if(j == 2)
				{//odd
					if(ints[i]%2==0)
						dp[i][j] = Math.min(dp[i][j], minPred + 1);
					else
						dp[i][j] = Math.min(dp[i][j], minPred);
				}
			}
		}
		//for(long[] in:dp)
			//System.out.println(Arrays.toString(in));
		System.out.println(dp[dp.length-1][4]);
	}
	
}
