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 lint;

const int MAX_N = 2e5 + 5;
const lint INF = 1e15;
int N;
lint A[MAX_N], dp[MAX_N][5];

int main(){
    ios::sync_with_stdio(false), cin.tie(nullptr);
    cin >> N;
    for(int i=1;i<=N;i++) cin >> A[i];

    for(int i=0;i<=N+1;i++) for(int j=0;j<5;j++) dp[i][j] = INF;
    dp[0][0] = 0;

    for(int i=1;i<=N+1;i++){
        dp[i][0] = dp[i-1][0] + A[i];
        dp[i][1] = min(dp[i-1][0], dp[i-1][1]) + A[i] % 2 + (A[i]==0 ? 2 : 0);
        for(int j=0;j<3;j++) dp[i][2] = min(dp[i][2], dp[i-1][j]+(A[i]+1)%2);
        for(int j=0;j<4;j++) dp[i][3] = min(dp[i][3], dp[i-1][j]+A[i]%2+(A[i]==0 ? 2 : 0));
        for(int j=0;j<5;j++) dp[i][4] = min(dp[i][4], dp[i-1][j]+A[i]);
    }

    cout << dp[N+1][4] << endl;
    return 0;
}
