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;
const long long infty = 9e18;
int cost (int a, int t) {
  if (t == 0 || t == 4) {
    return a;
  } else if (t == 1 || t == 3) {
    if (a == 0) {
      return 2;
    }
    return a & 1;
  } else {
    return (a + 1) & 1;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> dp(n + 1, vector<long long>(5, infty));
  for (int t = 0; t < 5; t++) {
    dp[0][t] = 0;
  }
  for (int i = 0; i < n; i++) {
    for (int t = 0; t < 5; t++) {
      dp[i + 1][t] = min(dp[i + 1][t], dp[i][t] + cost(a[i], t));
      if (t > 0) {
        dp[i + 1][t] = min(dp[i + 1][t], dp[i + 1][t - 1]);
      }
    }
  }
  long long ret = infty;
  for (int t = 0; t < 5; t++) {
    ret = min(ret, dp[n][t]);
  }
  cout << ret << '\n';
  return 0;
}