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<iostream>
#include<string>
#include<vector>
#include<iomanip>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
using namespace std;
#define ll long long
int main(){
    ll L,A;
    cin >> L;
    ll dp[L][5];
    cin >> A;
    dp[0][0]=A;
    if(A==0)dp[0][1]=2;else dp[0][1]=A%2;
    dp[0][2]=(A+1)%2;
    if(A==0)dp[0][3]=2;else dp[0][3]=A%2;
    dp[0][4]=A;
    ll tmp,i;
    for(i=1;i<L;i++){
      cin >> A;
      tmp =dp[i-1][0];
      dp[i][0]=tmp+A;
      tmp = min(tmp,dp[i-1][1]);
      if(A==0)dp[i][1]=tmp+2;else dp[i][1]=tmp+A%2;
      tmp = min(tmp,dp[i-1][2]);
      dp[i][2]=tmp+(A+1)%2;
      tmp = min(tmp,dp[i-1][3]);
      if(A==0)dp[i][3]=tmp+2;else dp[i][3]=tmp+A%2;
      tmp = min(tmp,dp[i-1][4]);
      dp[i][4]=tmp+A;
    }
    tmp=dp[L-1][0];
    for(i=1;i<5;i++){
      tmp=min(tmp,dp[L-1][i]);
    }
    cout << tmp;
    return 0;
}
