#include <iostream>
#include <algorithm>

using namespace std;

int main() {
    int l;
    cin >> l;
    long dp[l+1][5];
    fill(dp[0], dp[l+1], 0);
    for (int i=0; i<l; i++) {
        long a; cin >> a;
        dp[i+1][0] = dp[i][0] + a;
        dp[i+1][1] = min(dp[i][0], dp[i][1]) + a % 2 + (a == 0 ? 2 : 0);
        dp[i+1][2] = min({dp[i][0], dp[i][1], dp[i][2]}) + (a+1) % 2;
        dp[i+1][3] = min({dp[i][0], dp[i][1], dp[i][2], dp[i][3]}) + a % 2 + (a == 0 ? 2 : 0);
        dp[i+1][4] = min({dp[i][0], dp[i][1], dp[i][2], dp[i][3], dp[i][4]}) + a;
    }
    cout << min({dp[l][0], dp[l][1], dp[l][2], dp[l][3], dp[l][4]}) << '\n';
    return 0;
}
### Prompt

In Cpp, your task is to solve 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 <iostream>
#include <algorithm>

using namespace std;

int main() {
    int l;
    cin >> l;
    long dp[l+1][5];
    fill(dp[0], dp[l+1], 0);
    for (int i=0; i<l; i++) {
        long a; cin >> a;
        dp[i+1][0] = dp[i][0] + a;
        dp[i+1][1] = min(dp[i][0], dp[i][1]) + a % 2 + (a == 0 ? 2 : 0);
        dp[i+1][2] = min({dp[i][0], dp[i][1], dp[i][2]}) + (a+1) % 2;
        dp[i+1][3] = min({dp[i][0], dp[i][1], dp[i][2], dp[i][3]}) + a % 2 + (a == 0 ? 2 : 0);
        dp[i+1][4] = min({dp[i][0], dp[i][1], dp[i][2], dp[i][3], dp[i][4]}) + a;
    }
    cout << min({dp[l][0], dp[l][1], dp[l][2], dp[l][3], dp[l][4]}) << '\n';
    return 0;
}
```