#include <bits/stdc++.h>
using namespace std;
#define rep(i, m, n) for (int i = m; i < n; ++i)
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }

int main() {
  int L; cin >> L;
  vector<int> A(L);
  rep(i, 0, L) cin >> A[i];
  vector<vector<int>> dp(L + 1,vector<int>(5, 1e9));
  dp[0][0] = 0;
  rep(i, 0, L) {
    int pre = dp[i][0];
    rep(j, 0, 5) {
      chmin(pre, dp[i][j]);
      if(j == 0 || j == 4) {
        chmin(dp[i+1][j], pre + A[i]);
      } else if(j == 1 || j == 3) {
        int dif = A[i] == 0 ? 2 : A[i] % 2;
        chmin(dp[i+1][j], pre + dif);
      } else {
        chmin(dp[i+1][j], pre + (A[i] + 1) % 2);
      }
    }
  }
  int ans = 1e9;
  rep(i, 0, 5) chmin(ans, dp[L][i]);
  cout << ans << 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;
#define rep(i, m, n) for (int i = m; i < n; ++i)
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1; } return 0; }

int main() {
  int L; cin >> L;
  vector<int> A(L);
  rep(i, 0, L) cin >> A[i];
  vector<vector<int>> dp(L + 1,vector<int>(5, 1e9));
  dp[0][0] = 0;
  rep(i, 0, L) {
    int pre = dp[i][0];
    rep(j, 0, 5) {
      chmin(pre, dp[i][j]);
      if(j == 0 || j == 4) {
        chmin(dp[i+1][j], pre + A[i]);
      } else if(j == 1 || j == 3) {
        int dif = A[i] == 0 ? 2 : A[i] % 2;
        chmin(dp[i+1][j], pre + dif);
      } else {
        chmin(dp[i+1][j], pre + (A[i] + 1) % 2);
      }
    }
  }
  int ans = 1e9;
  rep(i, 0, 5) chmin(ans, dp[L][i]);
  cout << ans << endl;
  return 0;
}

```