#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

const int INF=1000100010;
int L;
int a[200001];
int dp[200001][5];
int main(){
    cin>>L;
    rep(i,L) cin>>a[i];
    rep(i,L+1)rep(j,5) dp[i][j]=INF;
    rep(i,5) dp[0][i]=0;
    rep(i,L){
        rep(j,5){
            rep(k,j+1){
                dp[i+1][j]=min(dp[i+1][j],dp[i][k]);
            }
            if(j==0||j==4) dp[i+1][j]+=a[i];
            else if(j==2) dp[i+1][j]+=(a[i]+1)%2;
            else if(a[i]==0) dp[i+1][j]+=2;
            else dp[i+1][j]+=a[i]%2;
        }
    }
    int ans=INF;
    rep(i,5) ans=min(ans,dp[L][i]);
    cout<<ans<<endl;
    return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

const int INF=1000100010;
int L;
int a[200001];
int dp[200001][5];
int main(){
    cin>>L;
    rep(i,L) cin>>a[i];
    rep(i,L+1)rep(j,5) dp[i][j]=INF;
    rep(i,5) dp[0][i]=0;
    rep(i,L){
        rep(j,5){
            rep(k,j+1){
                dp[i+1][j]=min(dp[i+1][j],dp[i][k]);
            }
            if(j==0||j==4) dp[i+1][j]+=a[i];
            else if(j==2) dp[i+1][j]+=(a[i]+1)%2;
            else if(a[i]==0) dp[i+1][j]+=2;
            else dp[i+1][j]+=a[i]%2;
        }
    }
    int ans=INF;
    rep(i,5) ans=min(ans,dp[L][i]);
    cout<<ans<<endl;
    return 0;
}
```