#include <iostream>
using namespace std;

const int MAXN = 2e5 + 10;
int dp[MAXN][5], n, a[MAXN];

int getval(int val, int t) {
	if (t == 0 || t == 4)
		return val;
	if (t == 1 || t == 3) {
		if (val == 0)
			return 2;
		return val % 2;
	}
	return !(val % 2);
}

int main() {
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for (int i = 1; i <= n; i++)
		for (int j = 0; j < 5; j++) {
			dp[i][j] = 1e9;
			for (int k = 0; k <= j; k++)
				dp[i][j] = min(dp[i][j], dp[i - 1][k] + getval(a[i], j));
		}
	cout << min(dp[n][0], min(dp[n][1], min(dp[n][2], min(dp[n][3], dp[n][4]))));
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <iostream>
using namespace std;

const int MAXN = 2e5 + 10;
int dp[MAXN][5], n, a[MAXN];

int getval(int val, int t) {
	if (t == 0 || t == 4)
		return val;
	if (t == 1 || t == 3) {
		if (val == 0)
			return 2;
		return val % 2;
	}
	return !(val % 2);
}

int main() {
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for (int i = 1; i <= n; i++)
		for (int j = 0; j < 5; j++) {
			dp[i][j] = 1e9;
			for (int k = 0; k <= j; k++)
				dp[i][j] = min(dp[i][j], dp[i - 1][k] + getval(a[i], j));
		}
	cout << min(dp[n][0], min(dp[n][1], min(dp[n][2], min(dp[n][3], dp[n][4]))));
}
```