#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)

int main()
{
ll n ;
cin >> n ;
vector<ll> a(n);
rep(i, n){
  cin >> a[i];
}
vector<vector<ll>> dp(n+1, vector<ll>(5,0) );
rep(i,n){
  dp[i+1][0]=dp[i][0]+a[i];
  ll r1=0;
  if(a[i]==0)r1=2;
  else r1=a[i]%2;
  dp[i+1][1]=min(dp[i][1]+r1,dp[i][0]+r1);
  ll r2=(a[i]+1)%2;
  dp[i+1][2]=min({dp[i][0]+r2,dp[i][1]+r2,dp[i][2]+r2});
  dp[i+1][3]=min({dp[i][0]+r1,dp[i][1]+r1,dp[i][2]+r1,dp[i][3]+r1});
  dp[i+1][4]=min({dp[i][0]+a[i],dp[i][1]+a[i],dp[i][2]+a[i],dp[i][3]+a[i],dp[i][4]+a[i]});
}
cout << *min_element(dp[n].begin(),dp[n].end()) << endl;
return 0;
}
### Prompt

Your task is to create 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;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)

int main()
{
ll n ;
cin >> n ;
vector<ll> a(n);
rep(i, n){
  cin >> a[i];
}
vector<vector<ll>> dp(n+1, vector<ll>(5,0) );
rep(i,n){
  dp[i+1][0]=dp[i][0]+a[i];
  ll r1=0;
  if(a[i]==0)r1=2;
  else r1=a[i]%2;
  dp[i+1][1]=min(dp[i][1]+r1,dp[i][0]+r1);
  ll r2=(a[i]+1)%2;
  dp[i+1][2]=min({dp[i][0]+r2,dp[i][1]+r2,dp[i][2]+r2});
  dp[i+1][3]=min({dp[i][0]+r1,dp[i][1]+r1,dp[i][2]+r1,dp[i][3]+r1});
  dp[i+1][4]=min({dp[i][0]+a[i],dp[i][1]+a[i],dp[i][2]+a[i],dp[i][3]+a[i],dp[i][4]+a[i]});
}
cout << *min_element(dp[n].begin(),dp[n].end()) << endl;
return 0;
}
```