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 <algorithm>
#include <vector>
#include <cstring>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <cstdio>
#include <cmath>
#include <limits.h>
using namespace std;
typedef long long LL;
const int N=2e5+10;

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

int main() {
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {
		scanf("%d",&a[i]);
	}
	memset(dp,-1,sizeof(dp));
	dp[0][0]=0;
	for(int i=0;i<n;i++) {
		for(int j=0;j<5;j++) {
			if(dp[i][j]==-1) continue;
			for(int k=j;k<5;k++) {
				int cost=0;
				if(k==0 || k==4) cost=a[i+1];
				else if((k==1 || k==3) && a[i+1]==0) cost=2;
				else if((k+a[i+1])%2==0) cost=1;
				else cost=0;

				LL x=cost+dp[i][j];
				if(dp[i+1][k]==-1 || x<dp[i+1][k]) dp[i+1][k]=x;
			}
		}
	}
	LL ans=-1;
	for(int i=0;i<5;i++) {
		if(ans==-1 || dp[n][i]<ans) ans=dp[n][i];
	}
	printf("%lld\n",ans);
}