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;

void chmin(int& a, int b){
    a = min(a, b);
}

int main(){
    int N, A[200000];
    cin >> N;
    for(int i=0; i<N; i++) cin >> A[i];

    static int dp[200001][5];
    for(int i=0; i<=N; i++) for(int j=0; j<=4; j++) dp[i][j] = 1e9;
    dp[0][0] = 0;

    for(int i=0; i<N; i++){
        int prv = dp[i][0];
        for(int j=0; j<=4; j++){
            chmin(prv, dp[i][j]);
            int diff;
            if(j==0 || j==4){
                diff = A[i];
            }else if(j==1 || j==3){
                diff = (A[i] == 0 ? 2 : A[i]%2);
            }else{
                diff = (A[i]+1)%2;
            }
            chmin(dp[i+1][j], prv + diff);
        }
    }

    int ans = 1e9;
    for(int j=0; j<=4; j++) chmin(ans, dp[N][j]);
    cout << ans << endl;
    return 0;
}