#include <bits/stdc++.h>

using namespace std;

int n;
long long f[5];

int main()
{
	cin>>n;
	int x;
	int i;
    for(i=1;i<=n;i++)
    {
    	cin>>x;
        long long minv=f[0]; 
		f[0]+=x;
        minv=min(minv,f[1]),f[1]=minv+(x?x&1:2);
        minv=min(minv,f[2]),f[2]=minv+(x?x+1&1:1);
        minv=min(minv,f[3]),f[3]=minv+(x?x&1:2);
        minv=min(minv,f[4]),f[4]=minv+x;
    }
    cout<<min(f[2],min(f[3],f[4]))<<endl;

    return 0;
}
### Prompt

Develop a solution in cpp 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;

int n;
long long f[5];

int main()
{
	cin>>n;
	int x;
	int i;
    for(i=1;i<=n;i++)
    {
    	cin>>x;
        long long minv=f[0]; 
		f[0]+=x;
        minv=min(minv,f[1]),f[1]=minv+(x?x&1:2);
        minv=min(minv,f[2]),f[2]=minv+(x?x+1&1:1);
        minv=min(minv,f[3]),f[3]=minv+(x?x&1:2);
        minv=min(minv,f[4]),f[4]=minv+x;
    }
    cout<<min(f[2],min(f[3],f[4]))<<endl;

    return 0;
}
```