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

int n,a[200000],sm1[200000],sm2[200000],x1[200000],x2[200000];

signed main(){
 
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>a[i],sm1[i]=a[i],sm2[i]=a[i];
  }

  for(int i=1;i<n;i++)sm1[i]+=sm1[i-1];
  for(int i=n-2;i>=0;i--)sm2[i]+=sm2[i+1];
  
  for(int i=n-1;i>=0;i--){
    int y=(i+1<n?x1[i+1]:0);
    if(a[i]==0)x1[i]=min(y+2,sm2[i]);
    else x1[i]=min(y+a[i]%2,sm2[i]);
  }

  for(int i=0;i<n;i++){
    int y=(i-1>=0?x2[i-1]:0);
    if(a[i]==0)x2[i]=min(y+2,sm1[i]);
    else x2[i]=min(y+a[i]%2,sm1[i]);
  }
  
  int ans=1e9,c=0;

  for(int i=0;i<n;i++){
    ans=min(ans,x1[i]+c);
    if(a[i]%2==0)c=min(c+1,x2[i]);
  }
  ans=min(ans,c);

  cout<<ans<<endl;
  return 0;
}