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 int;
#define long ll

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

    ll l;
    cin >> l;
    vector<ll> a(l), as(l+1);
    for(auto &e: a) cin >> e;

    partial_sum(a.begin(), a.end(), as.begin()+1);

    vector<ll> le(l+1), re(l+1);

    for(ll i=1; i<=l; i++) {
        ll x = a[i-1] > 0 ? a[i-1]&1 : 2;
        le[i] = min(le[i-1]+x, as[i]);
    }
    for(ll i=l-1; i>=0; i--) {
        ll x = a[i] > 0 ? a[i]&1 : 2;
        re[i] = min(re[i+1]+x, as[l] - as[i]);
    }
    vector<ll> rr(l+1);
    for(ll i=l-1; i>=0; i--) {
        ll x = a[i] > 0 ? ~a[i]&1 : 1;
        rr[i] = min(rr[i+1]+x, re[i]);
    }

    ll res = 1e18;
    for(ll i=0; i<=l; i++)
        res = min(res, le[i]+rr[i]);
    cout << res << endl;
}
