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;
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=1LL<<61;

int main() {
  ll n;
  cin >> n;
  vec a(n);
  mat dp(n+1,vec(5));
  for(ll i=0;i<n;i++) cin >> a[i];
  for(ll i=0;i<n;i++) {
    dp[i+1][0]=a[i]+dp[i][0];
    dp[i+1][1]=min(dp[i][0],dp[i][1]);
    dp[i+1][2]=min({dp[i][0],dp[i][1],dp[i][2]});
    dp[i+1][3]=min({dp[i][0],dp[i][1],dp[i][2],dp[i][3]});
    dp[i+1][4]=min({dp[i][0],dp[i][1],dp[i][2],dp[i][3],dp[i][4]})+a[i];
    if(a[i]%2) {
      dp[i+1][1]++;
      dp[i+1][3]++;
    }
    else {
      dp[i+1][2]++;
      if(a[i]==0) {
        dp[i+1][1]+=2;
        dp[i+1][3]+=2;
      }
    }
  }
  cout << *min_element(dp[n].begin(),dp[n].end()) << endl;
}