#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	int L; cin >> L;
	vector<LL> A(L);
	repp(i,0,L) cin >> A[i];
	vector<LL> dp(5,0);
	repp(i,0,L){
		vector<LL> nx(5,1e9+7);
		LL z = dp[0];
		nx[0] = z+A[i];
		z = min(z,dp[1]);
		nx[1] = z+(A[i]==0?2:(A[i]%2==0?0:1));
		z = min(z,dp[2]);
		nx[2] = z+(A[i]%2==1?0:1);
		z = min(z,dp[3]);
		nx[3] = z+(A[i]==0?2:(A[i]%2==0?0:1));
		z = min(z,dp[4]);
		nx[4] = z+A[i];
		swap(nx,dp);
	}
	repp(i,1,5) dp[0] = min(dp[0],dp[i]);
	cout << dp[0] << endl;
	return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

int main(){
	int L; cin >> L;
	vector<LL> A(L);
	repp(i,0,L) cin >> A[i];
	vector<LL> dp(5,0);
	repp(i,0,L){
		vector<LL> nx(5,1e9+7);
		LL z = dp[0];
		nx[0] = z+A[i];
		z = min(z,dp[1]);
		nx[1] = z+(A[i]==0?2:(A[i]%2==0?0:1));
		z = min(z,dp[2]);
		nx[2] = z+(A[i]%2==1?0:1);
		z = min(z,dp[3]);
		nx[3] = z+(A[i]==0?2:(A[i]%2==0?0:1));
		z = min(z,dp[4]);
		nx[4] = z+A[i];
		swap(nx,dp);
	}
	repp(i,1,5) dp[0] = min(dp[0],dp[i]);
	cout << dp[0] << endl;
	return 0;
}

```