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>
#include<iomanip>
#include<numeric>

using namespace std;
using ll = long long;
constexpr int mo = 1e9+7;
constexpr int  mod = mo;
constexpr ll inf = 1ll<<62;
vector<int> a;
ll cost(int i,int k){
    if(k == 0 or k== 4)return a[i];
    else if(k == 1 or k==3){
        if(a[i] == 0)return 2;
        if(a[i]%2)return 1;
    } else{
        if(a[i]%2 == 0)return 1;
    }
    return 0;
}
int main(){
    int l;
    cin>>l;
    a.resize(l);
    for(auto & b : a) cin>>b;
    vector<ll> from(5,inf);
    auto to = from;
    from[0] = 0;
    for(int i=0;i<l;++i){
        fill(to.begin(),to.end(),inf);
        for(int j=0;j<5;++j){
            for(int k=j;k<5;++k){
                to[k] = min(to[k],from[j] + cost(i,k));
            }
        }
        swap(to,from);
    }
    ll ret = inf;
    for(int i=0;i<5;++i){ret = min(ret,from[i]);}
    cout << ret << endl;
    return 0;
}

