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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll INF=1e18+7107;
int n,a[200005];
ll dpf[200005],dpb[200005],f[200005],sum[200005],ans;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<=n;i++) sum[i]=sum[i-1]+a[i];
	ans=INF;
	for(int i=1;i<=n;i++)
	{
		dpf[i]=min(sum[i-1],dpf[i-1])+(a[i]&1);
		if(!a[i]) dpf[i]+=2;
		//printf("dpf(%d)=(%lld,%lld)\n",i,dpf[i][0],dpf[i][1]);
	}
	for(int i=n;i>=1;i--)
	{
		dpb[i]=min(sum[n]-sum[i],dpb[i+1])+(a[i]&1);
		if(!a[i]) dpb[i]+=2;
		//printf("dfb(%d)=(%lld,%lld)\n",i,dpb[i][0],dpb[i][1]);
	}
	for(int i=1;i<=n;i++)
	{
		f[i]=min(min(sum[i-1],dpf[i-1]),f[i-1])+((a[i]&1)^1);
		ans=min(ans,min(f[i]+min(sum[n]-sum[i],dpb[i+1]),
			min(sum[i-1],dpf[i-1])+min(sum[n]-sum[i],dpb[i])));
		//printf("f(%d)=%lld\n",i,f[i]);
	}
	printf("%lld\n",ans);
	return 0;
}