#include <bits/stdc++.h>
#define int long long
using namespace std;

signed main(){
  int N;cin>>N;
  std::vector<int> v(N);
  for(int i=0;i<N;i++)cin>>v[i];
  std::vector<int> w(N);
  for(int i=0;i<N;i++)w[i]=v[i];
  for(int i=0;i<N;i++)w[i]=1-w[i]%2;
  std::vector<int> x(N);
  for(int i=0;i<N;i++)x[i]=v[i];
  for(int i=0;i<N;i++){
    if(x[i]==0)x[i]=2;
    else x[i]%=2;
  }
  std::vector<int> dp1(N+1),dp2(N+1),dp3(N+1),dp4(N+1),dp5(N+1);
  dp1[0]=dp2[0]=dp3[0]=dp4[0]=dp5[0]=0;
  for(int i=0;i<N;i++){
    dp1[i+1]=dp1[i]+v[i];
    dp2[i+1]=min(dp1[i]+x[i],dp2[i]+x[i]);
    dp3[i+1]=min(dp1[i]+w[i],min(dp2[i]+w[i],dp3[i]+w[i]));
    dp4[i+1]=min(min(dp1[i]+x[i],dp2[i]+x[i]),min(dp3[i]+x[i],dp4[i]+x[i]));
    dp5[i+1]=min(min(min(dp1[i]+v[i],dp2[i]+v[i]),dp3[i]+v[i]),min(dp4[i]+x[i],dp5[i]+v[i]));
  }
  int ans=min(dp1[N],min(dp2[N],min(dp3[N],min(dp4[N],dp5[N]))));
  cout<<ans<<endl;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int long long
using namespace std;

signed main(){
  int N;cin>>N;
  std::vector<int> v(N);
  for(int i=0;i<N;i++)cin>>v[i];
  std::vector<int> w(N);
  for(int i=0;i<N;i++)w[i]=v[i];
  for(int i=0;i<N;i++)w[i]=1-w[i]%2;
  std::vector<int> x(N);
  for(int i=0;i<N;i++)x[i]=v[i];
  for(int i=0;i<N;i++){
    if(x[i]==0)x[i]=2;
    else x[i]%=2;
  }
  std::vector<int> dp1(N+1),dp2(N+1),dp3(N+1),dp4(N+1),dp5(N+1);
  dp1[0]=dp2[0]=dp3[0]=dp4[0]=dp5[0]=0;
  for(int i=0;i<N;i++){
    dp1[i+1]=dp1[i]+v[i];
    dp2[i+1]=min(dp1[i]+x[i],dp2[i]+x[i]);
    dp3[i+1]=min(dp1[i]+w[i],min(dp2[i]+w[i],dp3[i]+w[i]));
    dp4[i+1]=min(min(dp1[i]+x[i],dp2[i]+x[i]),min(dp3[i]+x[i],dp4[i]+x[i]));
    dp5[i+1]=min(min(min(dp1[i]+v[i],dp2[i]+v[i]),dp3[i]+v[i]),min(dp4[i]+x[i],dp5[i]+v[i]));
  }
  int ans=min(dp1[N],min(dp2[N],min(dp3[N],min(dp4[N],dp5[N]))));
  cout<<ans<<endl;
}

```