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;

const int inf = 1e15 + 10, N = 3e5 + 10;
int a[N], n;
long long dp[N][6], ans = inf;

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];
		
	for (int i = 0; i < n; i++)
		for (int j = 0; j < 5; j++)
			dp[i][j] = inf;
		
	for (int i = 0; i < n; i++){
		for (int j = 0; j < 5; j++){
			int w = 0;
			if(a[i] == 0 && j != 0 && j != 4)
				w = 1 + (j % 2);
			else if(j == 2 && a[i] % 2 == 0)
				w = 1;
			else if((j == 1 || j == 3) && a[i] % 2 != 0)
				w = 1;
			else if((j == 0 || j == 4))
				w = a[i];
				
			for (int k = 0; k <= j; k++)
				dp[i][j] = min(dp[i - 1][k] + w, dp[i][j]);	
		}
	}
	
	for (int i = 0; i < 5; i++)
		ans = min(ans, dp[n - 1][i]);
	
	cout << ans;
}