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;
using ll=long long;
const ll MOD=1e9+7;
const int INF=1e9;
const ll LINF=(ll)1e18;

int main(){
    int l;
    cin>>l;
    int a[l];
    for(int i=0;i<l;i++){
        cin>>a[i];
    }
    ll dp[l+1][5];
    fill(dp[0],dp[l+1],LINF);
    dp[0][0]=0;
    for(int i=1;i<=l;i++){
        for(int j=0;j<5;j++){
            for(int k=0;k<=j;k++){
                if(j==0||j==4)dp[i][j]=min(dp[i][j],dp[i-1][k]+a[i-1]);
                else if(j==1||j==3)dp[i][j]=min(dp[i][j],dp[i-1][k]+(a[i-1]?a[i-1]%2:2LL));
                else dp[i][j]=min(dp[i][j],dp[i-1][k]+(a[i-1]+1)%2);
            }
        }
    }
    ll ans=LINF;
    for(int i=0;i<5;i++){
        ans=min(ans,dp[l][i]);
    }
    cout<<ans<<endl;
}
