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<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
	int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-1;
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';return x*f;
}

const int N = 200005;
int a[N];
LL f[N][5];

int main() {
	int n = read();
	for (int i = 1; i <= n; ++i) a[i] = read();
	for (int i = 1; i <= n; ++i) {
		f[i][0] = f[i - 1][0];
		for (int j = 1; j <= 4; ++j) f[i][j] = min(f[i - 1][j], f[i][j - 1]);
		f[i][0] += a[i];
		f[i][1] += (a[i] == 0) ? 2 : (a[i] & 1);
		f[i][2] += !(a[i] & 1);
		f[i][3] += (a[i] == 0) ? 2 : (a[i] & 1);
		f[i][4] += a[i];
	}
	LL ans = f[n][0];
	for (int i = 1; i <= 4; ++i) ans = min(ans, f[n][i]);
	cout << ans;
	return 0;
}