#include<bits/stdc++.h>
using namespace std;
const long long inf=1000000000000000000;
long long f[200010][5];
int n;
long long a[200010];
long long pan(int x)
{
	if (x==0) return 2; else if (x&1) return 1; else return 0;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	for (int i=0;i<=n;i++) 
	for (int j=0;j<5;j++) f[i][j]=inf;
	f[0][0]=0;
	for (int i=1;i<=n;i++) 
	{
		f[i][0]=f[i-1][0]+a[i];
	    f[i][1]=min(f[i-1][0],f[i-1][1])+pan(a[i]);
	    f[i][2]=min(min(f[i-1][0],f[i-1][1]),f[i-1][2])+(1-(a[i]%2));
	    f[i][3]=min(f[i-1][2],f[i-1][3])+pan(a[i]);
	    f[i][4]=min(min(f[i-1][2],f[i-1][3]),min(f[i-1][1],f[i-1][4]))+a[i];
	}
	cout << min(min(min(f[n][0],f[n][1]),min(f[n][2],f[n][3])),f[n][4]) << endl;
}
### 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<bits/stdc++.h>
using namespace std;
const long long inf=1000000000000000000;
long long f[200010][5];
int n;
long long a[200010];
long long pan(int x)
{
	if (x==0) return 2; else if (x&1) return 1; else return 0;
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
	}
	for (int i=0;i<=n;i++) 
	for (int j=0;j<5;j++) f[i][j]=inf;
	f[0][0]=0;
	for (int i=1;i<=n;i++) 
	{
		f[i][0]=f[i-1][0]+a[i];
	    f[i][1]=min(f[i-1][0],f[i-1][1])+pan(a[i]);
	    f[i][2]=min(min(f[i-1][0],f[i-1][1]),f[i-1][2])+(1-(a[i]%2));
	    f[i][3]=min(f[i-1][2],f[i-1][3])+pan(a[i]);
	    f[i][4]=min(min(f[i-1][2],f[i-1][3]),min(f[i-1][1],f[i-1][4]))+a[i];
	}
	cout << min(min(min(f[n][0],f[n][1]),min(f[n][2],f[n][3])),f[n][4]) << endl;
}
```