#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

typedef long long ll;

int main(){
    int l, a;
    cin >> l;
    vector<ll> dp(5, 0);
    for(int i = 0; i < l; i++){
        cin >> a;
        vector<ll> next(5);
        next[0] = dp[0] + a;
        next[1] = min(dp[0],dp[1]) + a%2 + 2*(a==0);
        next[2] = min({dp[0], dp[1], dp[2]}) + (1-a%2);
        next[3] = min({dp[0], dp[1], dp[2], dp[3]}) + a%2 + 2*(a==0);
        next[4] = *min_element(dp.begin(), dp.end()) + a;
        dp = next;
    }
    cout << *min_element(dp.begin(), dp.end()) << 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<vector>
#include<algorithm>
using namespace std;

typedef long long ll;

int main(){
    int l, a;
    cin >> l;
    vector<ll> dp(5, 0);
    for(int i = 0; i < l; i++){
        cin >> a;
        vector<ll> next(5);
        next[0] = dp[0] + a;
        next[1] = min(dp[0],dp[1]) + a%2 + 2*(a==0);
        next[2] = min({dp[0], dp[1], dp[2]}) + (1-a%2);
        next[3] = min({dp[0], dp[1], dp[2], dp[3]}) + a%2 + 2*(a==0);
        next[4] = *min_element(dp.begin(), dp.end()) + a;
        dp = next;
    }
    cout << *min_element(dp.begin(), dp.end()) << endl;
    return 0;
}
```