#include "bits/stdc++.h"
using namespace std;
signed main(){
    int n;
    cin >> n;
    long long x, sv = 0, sp = 0, sz = 0, min1 = 0, min2 = 0, min3 = 0, min4 = 0;
    for(int i = 1; i <= n; i++){
        cin >> x;
        sv+=x; sp+=x%2; sz+=!x;
        min4 = min(min4,
            (min3 = min(min3,
                (min2 = min(min2,
                    (min1 = min(min1,
                        sv-sp-2*sz
                    )) + 2*sz+2*sp-i
                )) - 2*sz-2*sp+i
            )) - sv+sp+2*sz
        );
    }
    cout << min4 + sv << endl;
}
### Prompt

Generate 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;
signed main(){
    int n;
    cin >> n;
    long long x, sv = 0, sp = 0, sz = 0, min1 = 0, min2 = 0, min3 = 0, min4 = 0;
    for(int i = 1; i <= n; i++){
        cin >> x;
        sv+=x; sp+=x%2; sz+=!x;
        min4 = min(min4,
            (min3 = min(min3,
                (min2 = min(min2,
                    (min1 = min(min1,
                        sv-sp-2*sz
                    )) + 2*sz+2*sp-i
                )) - 2*sz-2*sp+i
            )) - sv+sp+2*sz
        );
    }
    cout << min4 + sv << endl;
}
```