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;
typedef signed long long int ll;
#define INF (1<<29)
int l;
//ll a[200005];
ll a;
ll dp[200005][5]={0};
int main(){
    cin>>l;
    //for(int i=0;i<l;i++) cin>>a[i];
    
    /*for(int i=0;i<5;i++){
        for(int j=0;j<l+2;j++){
            dp[j][i]=INF;
        }
    }*/
    
    //for(int i=0;i<5;i++) dp[0][i]=0;
    
    for(int i=1;i<=l;i++){
        cin>>a;
        int b=(a==0?2:(a%2));
        
        int c=!(a%2);
        
        dp[i][0]=dp[i-1][0]+a;
        
        dp[i][1]=min(dp[i][0],dp[i-1][1]+b);
        
        dp[i][2]=min(dp[i][1],dp[i-1][2]+c);
        
        dp[i][3]=min(dp[i][2],dp[i-1][3]+b);
        
        dp[i][4]=min(dp[i][3],dp[i-1][4]+a);
        
    }
    cout<<dp[l][4]<<endl;
    return 0;
    
}
