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

int main() {

  ios::sync_with_stdio(false);
  cout.tie(nullptr);
  cin.tie(nullptr);

  int l;
  cin >> l;

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

  vector<vector<long long>> f(2, vector<long long> (l + 1, 0)), r = f;
  long long sum = 0, ans = l;
  for (int i = 1; i <= l; ++i) {
    sum += a[i];
    f[0][i] = min(sum, f[0][i - 1] + (a[i] & 1) + 2 * (a[i] == 0));
    f[1][i] = min(sum, min(f[0][i - 1], f[1][i - 1]) + !(a[i] & 1));
  }

  sum = 0;
  for (int i = l - 1; i >= 0; --i) {
    sum += a[i + 1];
    r[0][i] = min(sum, r[0][i + 1] + (a[i + 1] & 1) + 2 * (a[i + 1] == 0));
    r[1][i] = min(sum, min(r[0][i + 1], r[1][i + 1]) + !(a[i + 1] & 1));
    ans = min({ans, f[0][i] + r[0][i], f[0][i] + r[1][i], f[1][i] + r[0][i]});
  }

  cout << ans << '\n';
}


