#include <iostream>
#include <algorithm>

using namespace std;
using ll = long long;

const ll INF = 1LL << 50;

ll f(int k, ll a) {
    if (k == 2) {  // 奇数の区間
        return (a + 1) % 2;
    } else if (1 <= k && k <= 3) {  // 偶数の区間
        return (a == 0 ? 2 : a % 2);
    } else {  // 無の区間
        return a;
    }
}

int main() {
    int N;
    cin >> N;
    ll a[N + 2];
    for (int i = 1; i <= N; ++i) cin >> a[i];
    a[N + 1] = 0;

    ll dp[N + 2][5];
    fill(dp[0], dp[1], 0);
    fill(dp[1], dp[N + 2], INF);

    for (int i = 1; i <= N + 1; ++i) {
        for (int k = 0; k < 5; ++k) {
            ll m = *min_element(dp[i - 1], dp[i - 1] + k + 1);
            dp[i][k] = m + f(k, a[i]);
        }
    }

    cout << dp[N + 1][4] << endl;
    return 0;
}

### Prompt

Please create a solution in Cpp to 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 <iostream>
#include <algorithm>

using namespace std;
using ll = long long;

const ll INF = 1LL << 50;

ll f(int k, ll a) {
    if (k == 2) {  // 奇数の区間
        return (a + 1) % 2;
    } else if (1 <= k && k <= 3) {  // 偶数の区間
        return (a == 0 ? 2 : a % 2);
    } else {  // 無の区間
        return a;
    }
}

int main() {
    int N;
    cin >> N;
    ll a[N + 2];
    for (int i = 1; i <= N; ++i) cin >> a[i];
    a[N + 1] = 0;

    ll dp[N + 2][5];
    fill(dp[0], dp[1], 0);
    fill(dp[1], dp[N + 2], INF);

    for (int i = 1; i <= N + 1; ++i) {
        for (int k = 0; k < 5; ++k) {
            ll m = *min_element(dp[i - 1], dp[i - 1] + k + 1);
            dp[i][k] = m + f(k, a[i]);
        }
    }

    cout << dp[N + 1][4] << endl;
    return 0;
}

```