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;
#define int long long
const int INF = 1LL << 60;


int cost(int i, int s, const vector<int> &A){
    if(s == 0 || s == 4) return A[i];
    else if(s == 1 || s == 3){
        if(A[i] % 2 == 0 && A[i] > 0) return 0;
        else if(A[i] > 0) return 1;
        else return 2;
    }else{
        if(A[i] % 2 == 0) return 1;
        else return 0;
    }
}

signed main(){

    int L; cin >> L;
    vector<int> A(L);
    for(int i = 0; i < L; i++) cin >> A[i];

    vector<vector<int> > dp(L + 1, vector<int> (5, INF));
    dp[0][0] = 0;
    for(int i = 0; i < L; i++){
        for(int j = 0; j < 5; j++){
            for(int k = j; k < 5; k++){
                dp[i + 1][k] = min(dp[i + 1][k], dp[i][j] + cost(i, k, A));
            }
        }
    }

    int ans = INF;
    for(int j = 0; j < 5; j++) ans = min(ans, dp[L][j]);
    cout << ans << endl;
    return 0;
}