#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e15;

int main(){
  int l;
  cin >> l;
  ll a[l];
  rep(i,l) cin >> a[i];
  vector<vector<ll>> dp(l+1,vector<ll>(5,INF));
  dp[0][0] = 0;
  rep(i,l) {
    dp[i+1][0] = dp[i][0] + a[i];
    for(int j = 1; j <= 3; j++) {
      for(int k = 0; k <= j; k++) {
        if(j % 2) {
          if(a[i] == 0) dp[i+1][j] = min(dp[i+1][j],dp[i][k]+2);
          else dp[i+1][j] = min(dp[i+1][j],dp[i][k]+a[i]%2);
        }
        else dp[i+1][j] = min(dp[i+1][j],dp[i][k]+(1-a[i]%2));
      }
    }
    rep(j,4) dp[i+1][4] = min(dp[i+1][4],dp[i+1][j]); 
    dp[i+1][4] = min(dp[i+1][4],dp[i][4]+a[i]);
  }
  cout << dp[l][4] << endl;
  return 0;
}
### Prompt

Your challenge is to write 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;
#define rep(i,n) for(int i = 0; i < (int)n; i++)
using ll = long long;
const ll INF = 1e15;

int main(){
  int l;
  cin >> l;
  ll a[l];
  rep(i,l) cin >> a[i];
  vector<vector<ll>> dp(l+1,vector<ll>(5,INF));
  dp[0][0] = 0;
  rep(i,l) {
    dp[i+1][0] = dp[i][0] + a[i];
    for(int j = 1; j <= 3; j++) {
      for(int k = 0; k <= j; k++) {
        if(j % 2) {
          if(a[i] == 0) dp[i+1][j] = min(dp[i+1][j],dp[i][k]+2);
          else dp[i+1][j] = min(dp[i+1][j],dp[i][k]+a[i]%2);
        }
        else dp[i+1][j] = min(dp[i+1][j],dp[i][k]+(1-a[i]%2));
      }
    }
    rep(j,4) dp[i+1][4] = min(dp[i+1][4],dp[i+1][j]); 
    dp[i+1][4] = min(dp[i+1][4],dp[i][4]+a[i]);
  }
  cout << dp[l][4] << endl;
  return 0;
}
```