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

const int L = 2e5 + 5;
long long dp[L][5];

long long mdp(int i, int j) {
	if(i == 0) return 0;
	long long dr = dp[i][0];
	for(int ii = 1; ii <= j; ++ii)
		dr = min(dr, dp[i][ii]);
	return dr;
}

int l, a[L];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> l;
	for(int i = 1; i <= l; ++i) {
		cin >> a[i];
	}
	for(int i = 1; i <= l; ++i) {
		dp[i][0] = mdp(i - 1, 0) + a[i];
		dp[i][1] = mdp(i - 1, 1) + (a[i] == 0 ? 2 : (a[i] & 1));
		dp[i][2] = mdp(i - 1, 2) + (a[i] & 1 ^ 1);
		dp[i][3] = mdp(i - 1, 3) + (a[i] == 0 ? 2 : (a[i] & 1));
		dp[i][4] = mdp(i - 1, 4) + a[i];
	}
	cout << mdp(l, 4) << endl;
	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;

const int L = 2e5 + 5;
long long dp[L][5];

long long mdp(int i, int j) {
	if(i == 0) return 0;
	long long dr = dp[i][0];
	for(int ii = 1; ii <= j; ++ii)
		dr = min(dr, dp[i][ii]);
	return dr;
}

int l, a[L];

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin >> l;
	for(int i = 1; i <= l; ++i) {
		cin >> a[i];
	}
	for(int i = 1; i <= l; ++i) {
		dp[i][0] = mdp(i - 1, 0) + a[i];
		dp[i][1] = mdp(i - 1, 1) + (a[i] == 0 ? 2 : (a[i] & 1));
		dp[i][2] = mdp(i - 1, 2) + (a[i] & 1 ^ 1);
		dp[i][3] = mdp(i - 1, 3) + (a[i] == 0 ? 2 : (a[i] & 1));
		dp[i][4] = mdp(i - 1, 4) + a[i];
	}
	cout << mdp(l, 4) << endl;
	return 0;
}

```