#include <bits/stdc++.h>

using namespace std;

const int MX = 200000;
const long long INF = 1e15;

int cost(int a, int j) {
	if (j == 0 || j == 4) return a;
	if (j == 1 || j == 3) return a == 0 ? 2 : a % 2;
	if (j == 2) return (a + 1) % 2;
	assert(false);
}

long long dp[MX + 1][5];

int main() {
	int n;
	ignore = scanf("%d", &n);
	for (int i = 1; i <= n; i++) fill(dp[i], dp[i] + 5, INF);
	for (int i = 0, a; i < n; i++) {
		ignore = scanf("%d", &a);
		for (int j = 0; j < 5; j++) {
			dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + cost(a, j));
			if (j + 1 < 5) {
				dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + cost(a, j + 1));
				dp[i][j + 1] = min(dp[i][j + 1], dp[i][j]);
			}
		}
	}
	
	printf("%lld\n", *min_element(dp[n], dp[n] + 5));
	
	return 0;
}

### Prompt

Please create a solution in Cpp to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

const int MX = 200000;
const long long INF = 1e15;

int cost(int a, int j) {
	if (j == 0 || j == 4) return a;
	if (j == 1 || j == 3) return a == 0 ? 2 : a % 2;
	if (j == 2) return (a + 1) % 2;
	assert(false);
}

long long dp[MX + 1][5];

int main() {
	int n;
	ignore = scanf("%d", &n);
	for (int i = 1; i <= n; i++) fill(dp[i], dp[i] + 5, INF);
	for (int i = 0, a; i < n; i++) {
		ignore = scanf("%d", &a);
		for (int j = 0; j < 5; j++) {
			dp[i + 1][j] = min(dp[i + 1][j], dp[i][j] + cost(a, j));
			if (j + 1 < 5) {
				dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + cost(a, j + 1));
				dp[i][j + 1] = min(dp[i][j + 1], dp[i][j]);
			}
		}
	}
	
	printf("%lld\n", *min_element(dp[n], dp[n] + 5));
	
	return 0;
}

```