#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int T[5]={2,0,1,0,2};
int n,a[200005],c[3][200005];
LL dp[200005][5];
int main(){
	int i,j,k;
	scanf("%d",&n);
	for(i=0;i<n;++i){
		scanf("%d",a+i);
		if(!a[i]){
			c[1][i]=1;
			c[0][i]=2;
		}
		else if(a[i]&1){
			c[1][i]=0;
			c[0][i]=1;
		}
		else{
			c[1][i]=1;
			c[0][i]=0;
		}
		c[2][i]=a[i];
		for(j=0;j<5;++j){
			dp[i][j]=1000000000000000000ll;
		}
	}
	for(i=0;i<5;++i) dp[0][i]=(LL)c[T[i]][0];
	LL res=1000000000000000000ll;
	for(i=1;i<n;++i){
		for(j=0;j<5;++j){
			for(k=0;k<=j;++k){
				dp[i][j]=min(dp[i][j],dp[i-1][k]+(LL)c[T[j]][i]);
			}
		}
	}
	for(j=0;j<5;++j) res=min(res,dp[n-1][j]);
	printf("%lld\n",res);
	return 0;
}
### 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>
using namespace std;
typedef long long LL;
const int T[5]={2,0,1,0,2};
int n,a[200005],c[3][200005];
LL dp[200005][5];
int main(){
	int i,j,k;
	scanf("%d",&n);
	for(i=0;i<n;++i){
		scanf("%d",a+i);
		if(!a[i]){
			c[1][i]=1;
			c[0][i]=2;
		}
		else if(a[i]&1){
			c[1][i]=0;
			c[0][i]=1;
		}
		else{
			c[1][i]=1;
			c[0][i]=0;
		}
		c[2][i]=a[i];
		for(j=0;j<5;++j){
			dp[i][j]=1000000000000000000ll;
		}
	}
	for(i=0;i<5;++i) dp[0][i]=(LL)c[T[i]][0];
	LL res=1000000000000000000ll;
	for(i=1;i<n;++i){
		for(j=0;j<5;++j){
			for(k=0;k<=j;++k){
				dp[i][j]=min(dp[i][j],dp[i-1][k]+(LL)c[T[j]][i]);
			}
		}
	}
	for(j=0;j<5;++j) res=min(res,dp[n-1][j]);
	printf("%lld\n",res);
	return 0;
}
```