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;

int main(){
    long L;
    cin >> L;
    vector<long> A(L), C({0}), B(L + 1), D(L + 1);
    for(auto& i : A){
        cin >> i;
        C.push_back(1 ^ (i & 1));
    }
    partial_sum(C.begin(), C.end(), C.begin());

    long s(0);
    auto c = [](long a){return a == 0 ? 2 : a & 1;};
    for(int i = 1; i <= L; ++i){
        s += A[i - 1];
        B[i] = min(s, B[i - 1] + c(A[i - 1]));
    }
    s = 0;
    for(int i = L - 1; i >= 0; --i){
        s += A[i];
        D[i] = min(s, D[i + 1] + c(A[i]));
    }

    long mindex(0), ans(numeric_limits<long>::max() / 3);
    for(int i = 0; i <= L; ++i){
        if(B[mindex] - C[mindex] > B[i] - C[i])mindex = i;
        ans = min(ans, B[mindex] - C[mindex] + C[i] + D[i]);
    }

    cout << ans << endl;

    return 0;
}