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;

using ll = long long;

ll Solve() {
    ll l;
    cin >> l;

    vector<ll> a(l + 1);
    for (int i = 1; i <= l; ++i) cin >> a[i];

    constexpr int N = 5;
    vector<vector<ll>> dp(l + 1, vector<ll>(N, 1e11));
    dp[0][0] = 0;

    for (int i = 1; i <= l; ++i) {
        for (int j = 0; j < N; ++j) {
            for (int k = 0; k <= j; ++k) dp[i][j] = min(dp[i][j], dp[i - 1][k]);
            if (j == 0 || j == 4) dp[i][j] += a[i];
            else if (j == 1 || j == 3) dp[i][j] += (a[i] == 0 ? 2 : (a[i] % 2 == 1));
            else if (j == 2) dp[i][j] += (a[i] == 0 ? 1 : (a[i] % 2 == 0));
        }
    }

    return *min_element(dp[l].begin(), dp[l].end());
}

int main() {
    cin.tie(0); ios::sync_with_stdio(false);

    cout << Solve() << endl;

    return 0;
}
