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<cstdlib>
#include<cstring>
#include<algorithm>

typedef long long LL;

const int N=200005;
const LL inf=(LL)1e16;

int n,a[N];
LL s[N][3],dp[N][5];

void work()
{
	for (int i=1;i<=n;i++)
	{
		int x=a[i];
		if (!x) s[i][2]=2,s[i][1]=1,s[i][0]=0;
		else if (x&1) s[i][2]=1,s[i][1]=0,s[i][0]=x;
		else s[i][2]=0,s[i][1]=1,s[i][0]=x;
	}
	for (int i=1;i<=n+1;i++)
	{
		LL mn=dp[i-1][0];
		dp[i][0]=dp[i-1][0]+s[i][0];
		mn=std::min(mn,dp[i-1][1]);
		dp[i][1]=mn+s[i][2];
		mn=std::min(mn,dp[i-1][2]);
		dp[i][2]=mn+s[i][1];
		mn=std::min(mn,dp[i-1][3]);
		dp[i][3]=mn+s[i][2];
		mn=std::min(mn,dp[i-1][4]);
		dp[i][4]=mn+s[i][0];
	}
}

int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&a[i]);
	work();
	printf("%lld\n",dp[n+1][4]);
	return 0;
}