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
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int maxn = 200100;
const long long oo = 1e18;
int n;
long long a[maxn];
long long f[maxn][5];
int main()
{
	std::ios::sync_with_stdio(false);
	cin >> n;
	for(int i = 1; i <= n; i ++)
		cin >> a[i];
	for(int i = 1; i <= n; i ++)
		for(int j = 0; j < 5; j ++)
			f[i][j] = oo;
	for(int i = 1; i <= n; i ++)
	{
		long long mn = f[i - 1][0];
		f[i][0] = mn + a[i];
		mn = min(mn, f[i - 1][1]);
		f[i][1] = mn + (a[i] == 0 ? 2 : a[i] % 2);
		mn = min(mn, f[i - 1][2]);
		f[i][2] = mn + (a[i] % 2 != 1);
		mn = min(mn, f[i - 1][3]);
		f[i][3] = mn + (a[i] == 0 ? 2 : a[i] % 2);
		mn = min(mn, f[i - 1][4]);
		f[i][4] = mn + a[i];
	}
	cout << min(f[n][1], min(f[n][2], min(f[n][3], f[n][4]))) << endl;
	return 0;
}