#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    ll l,ans=1e9;
    cin >> l;
    vector<ll> a(l+1);
    for(int i = 0;i < l;i ++){
        cin >> a[i+1];
    }

    ll dp[200001][5];
    fill(dp[0],dp[200001],1e9);
    dp[0][0] = 0;

    for(int i = 1;i <= l;i ++){
        for(int j = 0;j < 5;j ++){
            ll m=1e9;
            for(int k = 0;k <= j;k ++){
                m = min(dp[i-1][k],m);
            }
            if(j==0||j==4) dp[i][j] = m+a[i];
            else if(j==1||j==3){
                dp[i][j] = m+a[i]%2;
                if(a[i]==0) dp[i][j] += 2;
            }
            else dp[i][j] = m+(a[i]+1)%2;
        }
    }
    for(int i = 0;i < 5;i ++){
        ans = min(ans,dp[l][i]);
    }
    cout << ans << endl;
    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;

int main(){
    ll l,ans=1e9;
    cin >> l;
    vector<ll> a(l+1);
    for(int i = 0;i < l;i ++){
        cin >> a[i+1];
    }

    ll dp[200001][5];
    fill(dp[0],dp[200001],1e9);
    dp[0][0] = 0;

    for(int i = 1;i <= l;i ++){
        for(int j = 0;j < 5;j ++){
            ll m=1e9;
            for(int k = 0;k <= j;k ++){
                m = min(dp[i-1][k],m);
            }
            if(j==0||j==4) dp[i][j] = m+a[i];
            else if(j==1||j==3){
                dp[i][j] = m+a[i]%2;
                if(a[i]==0) dp[i][j] += 2;
            }
            else dp[i][j] = m+(a[i]+1)%2;
        }
    }
    for(int i = 0;i < 5;i ++){
        ans = min(ans,dp[l][i]);
    }
    cout << ans << endl;
    return 0;
}
```