#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ll long long
ll gcd(int x, int y) { return (x % y)? gcd(y, x % y): y; }    //最大公約数
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }     //最小公倍数
using Graph = vector<vector<int>>;
ll inf=300000000000000000;
const double PI = 3.14159265358979323846;
int main(){
  int l;
  cin >> l;
  ll a[l];
  rep(i,l)cin >> a[i];
  ll b[l][5];
  rep(i,l){
    b[i][0]=a[i];
    b[i][4]=a[i];
    if(a[i]==0){
      b[i][1]=2;
      b[i][2]=1;
      b[i][3]=2;
    }
    else{
      b[i][1]=a[i]%2;
      b[i][3]=a[i]%2;
      b[i][2]=(a[i]+1)%2;
    }
  }
  ll dp[l+1][5];
  rep(i,l+1){
    rep(j,5)dp[i][j]=inf;
  }
  dp[0][0]=0;
  rep(i,l+1){
    rep(j,4)dp[i][j+1]=min(dp[i][j],dp[i][j+1]);
    if(i!=l){
      rep(j,5)dp[i+1][j]=dp[i][j]+b[i][j];
    }
  }
  cout << dp[l][4] << endl;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ll long long
ll gcd(int x, int y) { return (x % y)? gcd(y, x % y): y; }    //最大公約数
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }     //最小公倍数
using Graph = vector<vector<int>>;
ll inf=300000000000000000;
const double PI = 3.14159265358979323846;
int main(){
  int l;
  cin >> l;
  ll a[l];
  rep(i,l)cin >> a[i];
  ll b[l][5];
  rep(i,l){
    b[i][0]=a[i];
    b[i][4]=a[i];
    if(a[i]==0){
      b[i][1]=2;
      b[i][2]=1;
      b[i][3]=2;
    }
    else{
      b[i][1]=a[i]%2;
      b[i][3]=a[i]%2;
      b[i][2]=(a[i]+1)%2;
    }
  }
  ll dp[l+1][5];
  rep(i,l+1){
    rep(j,5)dp[i][j]=inf;
  }
  dp[0][0]=0;
  rep(i,l+1){
    rep(j,4)dp[i][j+1]=min(dp[i][j],dp[i][j+1]);
    if(i!=l){
      rep(j,5)dp[i+1][j]=dp[i][j]+b[i][j];
    }
  }
  cout << dp[l][4] << endl;
}

```