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"
using namespace std;
#define int long long
#define endl '\n'
int mod=1e9+7;

signed main(){
  int n;
  cin>>n;
  vector<int> a(n);
  for(int i=0;i<n;i++)cin>>a[i];
  vector<vector<int> > dp(n+1,vector<int>(5,1e18));
  dp[0][0]=0;
  dp[0][1]=0;
  dp[0][2]=0;
  dp[0][3]=0;
  dp[0][4]=0;

  for(int i=0;i<n;i++){
    int x=0;
    if(a[i]%2==1)x=1;
    if(a[i]==0)x=2;
    int y=0;
    if(a[i]%2==0)y=1;
    dp[i+1][0]=min(dp[i+1][0],dp[i][0]+a[i]);
    dp[i+1][1]=min(dp[i+1][1],min(dp[i][0],dp[i][1])+x);
    dp[i+1][2]=min(dp[i+1][2],min(dp[i][0],min(dp[i][1],dp[i][2]))+y);
    dp[i+1][3]=min(dp[i+1][3],min(dp[i][0],min(dp[i][1],min(dp[i][2],dp[i][3])))+x);
    dp[i+1][4]=min(dp[i+1][4],min(dp[i][0],min(dp[i][1],min(dp[i][2],min(dp[i][3],dp[i][4]))))+a[i]);
  }
  cout<<min(dp[n][0],min(dp[n][1],min(dp[n][2],min(dp[n][3],dp[n][4]))))<<endl;
}
