#include <bits/stdc++.h>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;
const int MAX_L=2e5;
const ll INF=1e18;
ll A[MAX_L],L;
ll dp[MAX_L+1][5];
int main(){
	cin>>L;
	REP(i,L){
		cin>>A[i];
	}
	REP(i,5)dp[L][i]=0;
	for(int n=L-1;n>=0;n--){
		for(int i=4;i>=0;i--){
			if(i==0 || i==4){
				dp[n][i]=A[n]+dp[n+1][i];
			}else if(i==2){
				dp[n][i]=(A[n]%2==1 ? 0 : 1)+dp[n+1][i];
			}else{
				ll num;
				if(A[n]==0){
					num=2;
				}else if(A[n]%2==0){
					num=0;
				}else{
					num=1;
				}
				dp[n][i]=num+dp[n+1][i];
			}
			if(i<4){
				dp[n][i]=min(dp[n][i],dp[n][i+1]);
			}
		}
	}
	cout<<dp[0][0]<<endl;
}
### Prompt

Generate 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>
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;
const int MAX_L=2e5;
const ll INF=1e18;
ll A[MAX_L],L;
ll dp[MAX_L+1][5];
int main(){
	cin>>L;
	REP(i,L){
		cin>>A[i];
	}
	REP(i,5)dp[L][i]=0;
	for(int n=L-1;n>=0;n--){
		for(int i=4;i>=0;i--){
			if(i==0 || i==4){
				dp[n][i]=A[n]+dp[n+1][i];
			}else if(i==2){
				dp[n][i]=(A[n]%2==1 ? 0 : 1)+dp[n+1][i];
			}else{
				ll num;
				if(A[n]==0){
					num=2;
				}else if(A[n]%2==0){
					num=0;
				}else{
					num=1;
				}
				dp[n][i]=num+dp[n+1][i];
			}
			if(i<4){
				dp[n][i]=min(dp[n][i],dp[n][i+1]);
			}
		}
	}
	cout<<dp[0][0]<<endl;
}
```