#include <bits/stdc++.h>

using namespace std;

int main()
{
    int n, a, s1, s2, s3, s4;
    cin >> n;
    long long Dp[n+1][5], M;
    for (int i = 0; i < 5; i++)
    {
        Dp[0][i] = 0;
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> a;
        Dp[i][0] = Dp[i-1][0] + a;
        s1 = a%2;
        if (a < 2)
        {
            s1 = 2-a;
        }
        M = Dp[i-1][0];
        Dp[i][1] = min(M, Dp[i-1][1]) + s1;
        s2 = (a+1) % 2;
        M = min(M, Dp[i-1][1]);
        Dp[i][2] = min(Dp[i-1][2], M) + s2;
        s3 = s1;
        M = min(M, Dp[i-1][2]);
        Dp[i][3] = min(Dp[i-1][3], M) + s3;
        M = min(M, Dp[i-1][3]);
        Dp[i][4] = min(Dp[i-1][4], M) + a;
    }
    M = Dp[n][0];
    for (int i = 1; i <= 4; i++)
    {
        M = min(Dp[n][i], M);
    }
    cout << M;
    return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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;

int main()
{
    int n, a, s1, s2, s3, s4;
    cin >> n;
    long long Dp[n+1][5], M;
    for (int i = 0; i < 5; i++)
    {
        Dp[0][i] = 0;
    }
    for (int i = 1; i <= n; i++)
    {
        cin >> a;
        Dp[i][0] = Dp[i-1][0] + a;
        s1 = a%2;
        if (a < 2)
        {
            s1 = 2-a;
        }
        M = Dp[i-1][0];
        Dp[i][1] = min(M, Dp[i-1][1]) + s1;
        s2 = (a+1) % 2;
        M = min(M, Dp[i-1][1]);
        Dp[i][2] = min(Dp[i-1][2], M) + s2;
        s3 = s1;
        M = min(M, Dp[i-1][2]);
        Dp[i][3] = min(Dp[i-1][3], M) + s3;
        M = min(M, Dp[i-1][3]);
        Dp[i][4] = min(Dp[i-1][4], M) + a;
    }
    M = Dp[n][0];
    for (int i = 1; i <= 4; i++)
    {
        M = min(Dp[n][i], M);
    }
    cout << M;
    return 0;
}

```