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<cstdio>
#include<algorithm>
using namespace std;
long n,a[200000],dp[5][200001],ans;
int main(){
    scanf("%ld",&n);
    for(int i=0;i<n;i++){
        scanf("%ld",&a[i]);
    }
    for(int i=0;i<n;i++){
        for(int j=0;j<5;j++){
            dp[j][i+1]=dp[0][i];
            for(int k=1;k<=j;k++){
                dp[j][i+1]=min(dp[j][i+1],dp[k][i]);
            }
        }
        dp[0][i+1]+=a[i];
        dp[4][i+1]+=a[i];
        dp[2][i+1]+=(a[i]+1)%2;
        if(a[i]==0){
            dp[1][i+1]+=2;
            dp[3][i+1]+=2;
        }
        else{
            dp[1][i+1]+=a[i]%2;
            dp[3][i+1]+=a[i]%2;
        }
    }
    ans=dp[0][n];
    for(int i=1;i<5;i++){
        ans=min(ans,dp[i][n]);
    }
    printf("%ld\n",ans);
}