#include <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
int n;

ll a[202020];
ll dp[202020][5];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
    }


    for(int i=1;i<=n;i++){
        dp[i][0] = dp[i-1][0] + a[i];
        dp[i][1] = min({dp[i][0],dp[i-1][1] + (a[i]==0?2:a[i]%2), dp[i-1][0] + a[i]%2});
        dp[i][2] = min({dp[i][1], dp[i-1][2] + (a[i]%2?0:1)});
        dp[i][3] = min({dp[i][2], min(dp[i-1][2],dp[i-1][3])+(a[i]==0?2:a[i]%2)});
        dp[i][4] = min({dp[i][3], dp[i-1][3] + (a[i]==0?2:a[i]%2), dp[i-1][4] + a[i]});
    }

    printf("%lld\n", dp[n][4]);
}

### Prompt

Please provide a CPP coded solution 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 <cstdio>
#include <algorithm>
using namespace std;
typedef long long ll;
int n;

ll a[202020];
ll dp[202020][5];
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lld",&a[i]);
    }


    for(int i=1;i<=n;i++){
        dp[i][0] = dp[i-1][0] + a[i];
        dp[i][1] = min({dp[i][0],dp[i-1][1] + (a[i]==0?2:a[i]%2), dp[i-1][0] + a[i]%2});
        dp[i][2] = min({dp[i][1], dp[i-1][2] + (a[i]%2?0:1)});
        dp[i][3] = min({dp[i][2], min(dp[i-1][2],dp[i-1][3])+(a[i]==0?2:a[i]%2)});
        dp[i][4] = min({dp[i][3], dp[i-1][3] + (a[i]==0?2:a[i]%2), dp[i-1][4] + a[i]});
    }

    printf("%lld\n", dp[n][4]);
}

```