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);
    for ( int i = 0; i < L; i++ ) {
        cin >> A[i];
    }
    ll INF = 1LL << 60;
    vector<vector<ll>> dp(L+1, vector<ll>(5, INF));
    for ( int i = 0; i < 5; i++ ) {
        dp[0][i] = 0;
    }
    for ( int i = 0; i < L; i++ ) {
        ll a = A[i];
        vector<ll> b(5);
        b[0] = b[4] = a;
        b[1] = b[3] = a > 0 ? a % 2 : 2;
        b[2] = (a+1) % 2;
        for ( int j = 0; j < 5; j++ ) {
            for ( int k = 0; k <= j; k++ ) {
                dp[i+1][j] = min(dp[i+1][j], dp[i][k] + b[j]);
            }
        }
    }
    ll ans = INF;
    for ( int i = 0; i < 5; i++ ) {
        ans = min(ans, dp[L][i]);
    }
    return ans;
}

int main() {
    auto ans = solve();
    cout << ans << "\n";
    return 0;
}