#include <cstdio>
#include <iostream>
#define MN 201000

int a[MN];
long long f[MN][3];
long long s[MN];

int main()
{
	int n; scanf("%d", &n);
	long long ans = 1e18;
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]), s[i] = s[i - 1] + a[i];
	for(int i = 1; i <= n; i++)
	{
		f[i][0] = std::min(f[i - 1][0], s[i - 1]) + (a[i] == 0 ? 2 : ((a[i] & 1) ? 1 : 0));
		f[i][1] = std::min(f[i - 1][0], std::min(f[i - 1][1], s[i - 1])) + (a[i] == 0 ? 1 : ((a[i] & 1) ? 0 : 1));
		f[i][2] = std::min(f[i - 1][1], f[i - 1][2]) + (a[i] == 0 ? 2 : ((a[i] & 1) ? 1 : 0));		
		ans = std::min(ans, std::min(std::min(f[i][0], f[i][1]), std::min(f[i][2], s[i])) + s[n] - s[i]);
	}
	printf("%lld\n", ans);
}
### Prompt

Create a solution in Cpp for 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 <cstdio>
#include <iostream>
#define MN 201000

int a[MN];
long long f[MN][3];
long long s[MN];

int main()
{
	int n; scanf("%d", &n);
	long long ans = 1e18;
	for(int i = 1; i <= n; i++) scanf("%d", &a[i]), s[i] = s[i - 1] + a[i];
	for(int i = 1; i <= n; i++)
	{
		f[i][0] = std::min(f[i - 1][0], s[i - 1]) + (a[i] == 0 ? 2 : ((a[i] & 1) ? 1 : 0));
		f[i][1] = std::min(f[i - 1][0], std::min(f[i - 1][1], s[i - 1])) + (a[i] == 0 ? 1 : ((a[i] & 1) ? 0 : 1));
		f[i][2] = std::min(f[i - 1][1], f[i - 1][2]) + (a[i] == 0 ? 2 : ((a[i] & 1) ? 1 : 0));		
		ans = std::min(ans, std::min(std::min(f[i][0], f[i][1]), std::min(f[i][2], s[i])) + s[n] - s[i]);
	}
	printf("%lld\n", ans);
}
```