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<cstdio>
#include<string>
#include<cstring> 
#include<cassert>
#include<cmath>
#include<sstream>
#include<fstream>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<bitset>
using namespace std;
long long a[200005];
long long dp[200005][5];
int main(){
	int n;
	cin>>n;
	for(int i=1;i<=n;i++){
		cin>>a[i];
	}
	for(int i=1;i<=n;i++){
		dp[i][0]=dp[i-1][0]+a[i];
		dp[i][1]=min(dp[i-1][0],dp[i-1][1])+((a[i]==0)?2:(a[i]%2));
		dp[i][2]=min(dp[i-1][1],dp[i-1][2])+((a[i]==0)?1:(a[i]+1)%2);
		dp[i][3]=min(dp[i-1][2],dp[i-1][3])+((a[i]==0)?2:(a[i]%2));
		dp[i][4]=min(dp[i-1][3],dp[i-1][4])+a[i];
		for(int j=1;j<=4;j++) dp[i][j]=min(dp[i][j],dp[i][j-1]);
	}
	cout<<dp[n][4];
}