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;
#define modulo 1000000007
#define mod(mod_x) ((((long long)mod_x+modulo))%modulo)
#define Inf 1000000000000000



int main(){
	
	int N;
	cin>>N;
	
	vector<vector<long long>> dp(N+1,vector<long long> (5,Inf));
	
	for(int i=0;i<=N;i++){
		long long A;
		if(i!=0)cin>>A;
		for(int j=0;j<5;j++){
			if(i==0){
				dp[i][j]=0;
			}
			else{
				long long cost=0;
				if(j==0||j==4)cost=A;
				if(j==1||j==3){
					if(A%2==1)cost=1;
					if(A==0)cost=2;
				}
				if(j==2){
					if(A%2==0)cost=1;
				}
				for(int k=0;k<=j;k++){
					dp[i][j] = min(dp[i][j],dp[i-1][k]+cost);
				}
			}
		}
	}
	
	long long ans = Inf;
	/*
	for(int i=0;i<=N;i++){
		for(int j=0;j<5;j++){
			cout<<dp[i][j]<<',';
		}
		cout<<endl;
	}*/
	
	for(int i=0;i<5;i++){
		ans = min(ans,dp[N][i]);
	}
	cout<<ans<<endl;
		
	
	return 0;
	
}
