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;

int main() {
    int n;
    cin >> n;
    vector<int> as(n);
    long long sum = 0;
    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        as[i] = a;
        sum += a;
    }
    vector<vector<long long>> dp(n + 1, vector<long long>(5));
    for (int i = 0; i < n; i++) {
        dp[i + 1][0] = dp[i][0] + as[i];
        dp[i + 1][1] = min(min(dp[i][0], dp[i][1]) + (as[i] ? as[i] & 1 : 2), dp[i + 1][0]);
        dp[i + 1][2] = min(min(dp[i][1], dp[i][2]) + (as[i] & 1 ? 0 : 1), dp[i + 1][1]);
        dp[i + 1][3] = min(min(dp[i][2], dp[i][3]) + (as[i] ? as[i] & 1 : 2), dp[i + 1][2]);
        dp[i + 1][4] = min(min(dp[i][3], dp[i][4]) + as[i], dp[i + 1][3]);
    }
    cout << dp[n][4] << endl;
}
