#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int Maxn = 200005;

int n;
int A[Maxn];
ll zer[Maxn], ev[Maxn], od[Maxn];
ll res;

int evCost(int a) { return a == 0? 2: bool(a % 2 != 0); }

int odCost(int a) { return bool(a % 2 != 1); }

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &A[i]);
    for (int i = n - 1; i >= 0; i--) {
        zer[i] = A[i + 1] + zer[i + 1];
        ev[i] = min(zer[i], evCost(A[i + 1]) + ev[i + 1]);
        od[i] = min(ev[i], odCost(A[i + 1]) + od[i + 1]);
    }
    res = od[0];
    ll myzer = 0, myev = 0;
    for (int i = 1; i <= n; i++) {
        myzer = A[i] + myzer;
        myev = min(myzer, evCost(A[i]) + myev);
        res = min(res, myev + od[i]);
    }
    cout << res << endl;
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int Maxn = 200005;

int n;
int A[Maxn];
ll zer[Maxn], ev[Maxn], od[Maxn];
ll res;

int evCost(int a) { return a == 0? 2: bool(a % 2 != 0); }

int odCost(int a) { return bool(a % 2 != 1); }

int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &A[i]);
    for (int i = n - 1; i >= 0; i--) {
        zer[i] = A[i + 1] + zer[i + 1];
        ev[i] = min(zer[i], evCost(A[i + 1]) + ev[i + 1]);
        od[i] = min(ev[i], odCost(A[i + 1]) + od[i + 1]);
    }
    res = od[0];
    ll myzer = 0, myev = 0;
    for (int i = 1; i <= n; i++) {
        myzer = A[i] + myzer;
        myev = min(myzer, evCost(A[i]) + myev);
        res = min(res, myev + od[i]);
    }
    cout << res << endl;
    return 0;
}

```