Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4
1
0
2
3

Sample Output 1

1

Assume that Snuke walks as follows:

He starts walking at coordinate 3 and finishes walking at coordinate 4, visiting coordinates 3,4,3,2,3,4 in this order.

Then, Snuke's four ears will contain 0,0,2,3 stones, respectively.
Ringo can satisfy the requirement by putting one stone in the first ear.

Sample Input 2

8
2
0
0
2
1
3
4
1

Sample Output 2

3

Sample Input 3

7
314159265
358979323
846264338
327950288
419716939
937510582
0

Sample Output 3

1

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

//#include "pch.h"
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

#define llong long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000//10^12

int main(void)
{
	//変数の宣言
	llong int l;
	static llong int a[200010];
	static llong int even[200010];
	static llong int odd[200010];
	static llong int sum[200010];
	static llong int left1[200010];
	static llong int left2[200010];
	static llong int left3[200010];
	static llong int left4[200010];
	llong int temp;


	llong int ans;
	llong int i;
	llong int flag;

	//データの読み込み

	scanf("%lld", &l);
//	scanf_s("%lld", &l);


	for (i = 0;i < l;i++) {
		scanf("%lld", &a[i]);
	}
//	for (i = 0;i < l;i++) {
//		scanf_s("%lld", &a[i]);
//	}

	//	printf("nは%dです\n", n);
	//	printf("データの読み込み終了\n");
		//実際の処理

		//累積和even[k]を作る
	for (i = 0;i < l;i++) {
		if (a[i] < 2) {
			even[i] = 2 - a[i];
		}
		else {
			even[i] = a[i] % 2;
		}
	}
	//oddを作る
	for (i = 0;i < l;i++) {
		odd[i] = (a[i] + 1) % 2;
		
	}

	//sumを作る
	for (i = 0;i < l;i++) {
		if (i == 0) {
			sum[i] = a[i];
		}
		else {
			sum[i] = sum[i - 1] + a[i];
		}
	}

	for (i = 0;i < l;i++) {
		if (i == 0) {
			left1[i] = min(even[0], a[0]);
		}else{
			left1[i] = min(left1[i - 1] + even[i], sum[i]);
		}
	}

	for (i = 0;i < l;i++) {
		if (i == 0) {
			left2[i] = min(left1[0], odd[0]);
		}
		else {
			left2[i] = min(left2[i - 1] + odd[i], left1[i]);
		}
	}

	for (i = 0;i < l;i++) {
		if (i == 0) {
			left3[i] = min(left2[0], even[0]);
		}
		else {
			left3[i] = min(left3[i - 1] + even[i], left2[i]);
		}
	}

	for (i = 0;i < l;i++) {
		if (i == 0) {
			left4[i] = min(left3[0], sum[0]);
		}
		else {
			left4[i] = min(left4[i - 1] + a[i], left3[i]);
		}
	}

	//	printf("計算部分終了\n");

		//出力
	printf("%lld", left4[l-1]);
/*
	for (i = 0;i < l;i++) {
		printf("i=%lld,left1[i]=%lld\n", i, left1[i]);
	}

	for (i = 0;i < l;i++) {
		printf("i=%lld,left2[i]=%lld\n", i, left2[i]);
	}

	for (i = 0;i < l;i++) {
		printf("i=%lld,left3[i]=%lld\n", i, left3[i]);
	}

	for (i = 0;i < l;i++) {
		printf("i=%lld,left4[i]=%lld\n", i, left4[i]);
	}
*/
	//	printf("結果の出力終了\n");

	return 0;
}