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 <cstdio>
#include <cstring>
#include <vector>
#include <tuple>

using namespace std;

int f1(int a) {
	if(a<2) {
		return 2-a;
	}
	return a & 1;
}
int f2(int a) {
	return (a ^ 1) & 1;
}

int main() {
	int N;
	scanf("%d", &N);
	long long st0 = 0;
	long long st1 = 0;
	long long st2 = 0;
	long long st3 = 0;
	long long st4 = 0;
	for(int n=0; n<N; ++n) {
		int a;
		scanf("%d", &a);
		long long bSt0 = st0;
		long long bSt1 = st1;
		long long bSt2 = st2;
		long long bSt3 = st3;
		long long bSt4 = st4;
		st0 = bSt0 + a;
		st1 = min(bSt0, bSt1) + f1(a);
		st2 = min(min(bSt0, bSt1), bSt2) + f2(a);
		st3 = min(min(bSt0, bSt1), min(bSt2, bSt3)) + f1(a);
		st4 = min(min(min(bSt0, bSt1), min(bSt2, bSt3)), bSt4) + a;
	}
	long long ret = min(min(min(st0, st1), min(st2, st3)), st4);
	printf("%lld\n", ret);

	return 0;
}
