#include<bits/stdc++.h>
#define mx 100005
using namespace std;

long long l, a, dp[5];
int main(){
  scanf("%lld", &l);
  for(int i = 0; i < l; i++){
    scanf("%lld", &a);
    dp[0] += a;
    dp[1] = min(dp[1] + (a ? (a&1) : 2), dp[0]);
    dp[2] = min(dp[2] + ((a+1)&1), dp[1]);
    dp[3] = min(dp[3] + (a ? (a&1) : 2), dp[2]);
    dp[4] = min(dp[4] + a, dp[3]);
  }
  printf("%lld\n", dp[4]);
}
### Prompt

Your challenge is to write 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>
#define mx 100005
using namespace std;

long long l, a, dp[5];
int main(){
  scanf("%lld", &l);
  for(int i = 0; i < l; i++){
    scanf("%lld", &a);
    dp[0] += a;
    dp[1] = min(dp[1] + (a ? (a&1) : 2), dp[0]);
    dp[2] = min(dp[2] + ((a+1)&1), dp[1]);
    dp[3] = min(dp[3] + (a ? (a&1) : 2), dp[2]);
    dp[4] = min(dp[4] + a, dp[3]);
  }
  printf("%lld\n", dp[4]);
}
```