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 inf 0x3f3f3f3f
typedef long long ll;
const int N=2e5+5;
int n,a[N][3];
ll dp[N][5];
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i][0]);
        a[i][1]=a[i][0]%2==0;
        a[i][2]=a[i][0]==0?2:a[i][0]%2;
    }
    for(int i=1;i<=n;i++)
    {
        ll mn=1e18;
        for(int j=0;j<5;j++)
        {
            mn=min(mn,dp[i-1][j]);
            ll s;
            if(j==0) s=a[i][0];
            else if(j==1) s=a[i][2];
            else if(j==2) s=a[i][1];
            else if(j==3) s=a[i][2];
            else s=a[i][0];
            dp[i][j]=mn+s;
        }
    }
    ll ans=1e18;
    for(int i=0;i<5;i++) ans=min(ans,dp[n][i]);
    printf("%lld\n",ans);
}
