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
#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
signed 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, 1LL << 60));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    long long cur = 1LL << 60;
    for (int j = 0; j < 5; j++) {
      cur = min(cur, dp[i][j]);
      long long cost;
      if (j == 0 || j == 4) cost = a[i];
      if (j == 1 || j == 3) cost = (a[i] > 0 ? a[i] % 2 : 2);
      if (j == 2) cost = (a[i] % 2 == 0);
      dp[i + 1][j] = min(dp[i + 1][j], cur + cost);
    }
  }
  long long ans = 1LL << 60;
  for (int i = 0; i < 5; i++) ans = min(ans, dp[n][i]);
  cout << ans << endl;
  return 0;
}