#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long

int L;
ll dp[5];
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> L;
    register ll A;

    for(int i = 0; i < L; ++i){
        cin >> A;
        ll mn = dp[0];
        dp[0] += A;
        mn = min(mn, dp[1]), dp[1] = mn + (A == 0 ? 2 : A % 2);
        mn = min(mn, dp[2]), dp[2] = mn + (A + 1) % 2;
        mn = min(mn, dp[3]), dp[3] = mn + (A == 0 ? 2 : A % 2);
        mn = min(mn, dp[4]), dp[4] = mn + A;
    }
    cout << min({dp[0], dp[1], dp[2], dp[3], dp[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<iostream>
#include<algorithm>
using namespace std;
#define ll long long

int L;
ll dp[5];
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> L;
    register ll A;

    for(int i = 0; i < L; ++i){
        cin >> A;
        ll mn = dp[0];
        dp[0] += A;
        mn = min(mn, dp[1]), dp[1] = mn + (A == 0 ? 2 : A % 2);
        mn = min(mn, dp[2]), dp[2] = mn + (A + 1) % 2;
        mn = min(mn, dp[3]), dp[3] = mn + (A == 0 ? 2 : A % 2);
        mn = min(mn, dp[4]), dp[4] = mn + A;
    }
    cout << min({dp[0], dp[1], dp[2], dp[3], dp[4]}) << endl;
    return 0;
}
```