#include <bits/stdc++.h>
using namespace std;
int main() {
	int L;
	cin >> L;
	vector<long long> A(L);
	for (auto& x : A) cin >> x;

	vector<vector<long long>> a(L+1, vector<long long>(5));
	for (int i = 0; i < L; ++ i) {
		auto m = 1LL<<60;
		for (int j = 0; j < 5; ++ j) {
			m = min(m, a[i][j]);
			if (j == 0 || j == 4) {
				a[i+1][j] = m + A[i];
			} else {
				auto k = (A[i] + j + 1) % 2;
				if (A[i] == 0 && j%2 == 1) k = 2;
				a[i+1][j] = m + k;
			}
		}
	}
	/*for (int i = 0; i < L+1; ++ i) {
		for (int j = 0; j < 5; ++ j) {
			cerr << a[i][j] << " ";
		}
		cerr << endl;
	}*/
	auto r = 1LL<<60;
	for (int j = 0; j < 5; ++ j) r = min(r, a[L][j]);
	cout << r << endl;
}

### 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;
int main() {
	int L;
	cin >> L;
	vector<long long> A(L);
	for (auto& x : A) cin >> x;

	vector<vector<long long>> a(L+1, vector<long long>(5));
	for (int i = 0; i < L; ++ i) {
		auto m = 1LL<<60;
		for (int j = 0; j < 5; ++ j) {
			m = min(m, a[i][j]);
			if (j == 0 || j == 4) {
				a[i+1][j] = m + A[i];
			} else {
				auto k = (A[i] + j + 1) % 2;
				if (A[i] == 0 && j%2 == 1) k = 2;
				a[i+1][j] = m + k;
			}
		}
	}
	/*for (int i = 0; i < L+1; ++ i) {
		for (int j = 0; j < 5; ++ j) {
			cerr << a[i][j] << " ";
		}
		cerr << endl;
	}*/
	auto r = 1LL<<60;
	for (int j = 0; j < 5; ++ j) r = min(r, a[L][j]);
	cout << r << endl;
}

```