#include <bits/stdc++.h>
#define INF 1000000007
using namespace std;
int n,a[200010],b[200010];
int dp[200010][5];
int k[]={0,2,1,2,0};
int main(){
	cin >> n;
	for(int i=1;i<=n;i++){
		cin >> a[i];
		b[i]=a[i];
		if(a[i]==0) continue;
		else if(a[i]%2==0) a[i]=2;
		else a[i]=1;
	}
	for(int i=1;i<=n;i++){
		for(int j=0;j<5;j++){
			dp[i][j]=INF;
			for(int l=0;l<=j;l++){
				dp[i][j]=min(dp[i][j],dp[i-1][l]);
			}
			if(a[i]==0) dp[i][j]+=k[j];
			else if(k[j]==0) dp[i][j]+=b[i];
			else dp[i][j]+=abs(a[i]-k[j]);
		}
	}
	int ans=INF;
	for(int i=0;i<5;i++) ans=min(ans,dp[n][i]);
	cout << ans << endl;
	return 0;
}
### Prompt

Your task is to create a CPP solution to the following problem:

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>
#define INF 1000000007
using namespace std;
int n,a[200010],b[200010];
int dp[200010][5];
int k[]={0,2,1,2,0};
int main(){
	cin >> n;
	for(int i=1;i<=n;i++){
		cin >> a[i];
		b[i]=a[i];
		if(a[i]==0) continue;
		else if(a[i]%2==0) a[i]=2;
		else a[i]=1;
	}
	for(int i=1;i<=n;i++){
		for(int j=0;j<5;j++){
			dp[i][j]=INF;
			for(int l=0;l<=j;l++){
				dp[i][j]=min(dp[i][j],dp[i-1][l]);
			}
			if(a[i]==0) dp[i][j]+=k[j];
			else if(k[j]==0) dp[i][j]+=b[i];
			else dp[i][j]+=abs(a[i]-k[j]);
		}
	}
	int ans=INF;
	for(int i=0;i<5;i++) ans=min(ans,dp[n][i]);
	cout << ans << endl;
	return 0;
}
```