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 inf = 1e18;
int n;
long a[200000];
long dp[5][200001];
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  
  for (int i = 0; i <= n; ++i) {
    for (int k = 0; k < 5; ++k) {
      dp[k][i] = inf;
    }
  }
  
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    for (int k = 0; k < 5; ++k) {
      for (int nk = k; nk < 5; ++nk) {
        long c;
        if (nk == 0 || nk == 4) c = a[i];
        if (nk == 1 || nk == 3) c = a[i] == 0 ? 2 : a[i] % 2;
        if (nk == 2) c = a[i] % 2 == 0;
        dp[nk][i + 1] = min(dp[nk][i + 1], dp[k][i] + c);
      }
    }
  }
  
  long mn = inf;
  for (int k = 0; k < 5; ++k) {
    mn = min(mn, dp[k][n]);
  }
  cout << mn << endl;
}

