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>
#define ll long long
using namespace std;
ll n,a[200005],f[200005][3];
int main()
{
	scanf("%lld",&n);
	memset(f,0x3f,sizeof(f));
	f[0][0]=0;
	ll sum=0;
	for(int i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		f[i][0]=min(f[i-1][0],sum)+(a[i]&1?1:(a[i]?0:2));
		f[i][1]=min(f[i-1][1],f[i-1][2])+(a[i]&1?1:(a[i]?0:2));
		f[i][2]=min(f[i-1][0],min(f[i-1][2],sum))+(a[i]&1?0:1);
//		printf("%lld %lld %lld\n",f[i][0],f[i][1],f[i][2]);
		sum+=a[i];
	}
	sum=0;
	ll ans=1000000000000000000ll;
	for(int i=n;i>=0;i--)
	{
		ans=min(ans,min(min(f[i][0],f[i][1]),f[i][2])+sum);
		sum+=a[i];
	}
	printf("%lld\n",ans);
	return 0;
}