#include <bits/stdc++.h>

using namespace std;

int l;
long long x[200005][5];
long long ta;

int main() {
	scanf("%d", &l);
	for (int i = 0; i < l; i++) {
		scanf("%lld", &ta);
		x[i + 1][0] = x[i][0] + ta;
		x[i + 1][1] = min(x[i + 1][0], x[i][1] + (ta == 0 ? 2 : (ta & 1)));
		x[i + 1][2] = min(x[i + 1][1], x[i][2] + ((ta & 1) ^ 1));
		x[i + 1][3] = min(x[i + 1][2], x[i][3] + (ta == 0 ? 2 : (ta & 1)));
		x[i + 1][4] = min(x[i + 1][3], x[i][4] + ta);
	}
	printf("%lld\n", x[l][4]);
	return 0;
}
### Prompt

Please formulate a CPP solution 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;

int l;
long long x[200005][5];
long long ta;

int main() {
	scanf("%d", &l);
	for (int i = 0; i < l; i++) {
		scanf("%lld", &ta);
		x[i + 1][0] = x[i][0] + ta;
		x[i + 1][1] = min(x[i + 1][0], x[i][1] + (ta == 0 ? 2 : (ta & 1)));
		x[i + 1][2] = min(x[i + 1][1], x[i][2] + ((ta & 1) ^ 1));
		x[i + 1][3] = min(x[i + 1][2], x[i][3] + (ta == 0 ? 2 : (ta & 1)));
		x[i + 1][4] = min(x[i + 1][3], x[i][4] + ta);
	}
	printf("%lld\n", x[l][4]);
	return 0;
}
```