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;

long long nos[200010];
long long dp[200010][4];
long long cuml[200010];
// 0 -> Ignore, 1 -> Odd, 2 -> Event
long long cost(int x, int t){
	if(t == 0) return x;
	else if(t == 1){
		if(x == 0) return 1;
		else return 1 - (x%2);
	}else{
		if(x == 0) return 2;
		else return x%2;
	}
}
int main(){
	int N; cin>>N;
	for(int i=1;i<=N;i++) cin>>nos[i];
	cuml[N] = nos[N];
	for(int i=N-1;i>=0;i--) cuml[i] = nos[i] + cuml[i+1];
	long long mcost = 1LL<<60;
	for(int i=1;i<=N;i++){
		dp[i][0] = dp[i-1][0] + cost(nos[i],0);
		dp[i][1] = min(dp[i][0], dp[i-1][1] + cost(nos[i],2)); // Even end meets start end
		dp[i][2] = min(dp[i][1], dp[i-1][2] + cost(nos[i],1)); // Odd end meets even end
		dp[i][3] = min(dp[i][2], dp[i-1][3] + cost(nos[i],2));
		mcost = min(mcost, dp[i][3] + cuml[i+1]);
	}
	cout<<mcost<<endl;
	return 0;
}