// ※※※ 解答不能 ※※※
// SovietPower氏.
// https://atcoder.jp/contests/yahoo-procon2019-qual/submissions/4241327
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define rep(i, n) for(int i = 0; i < n; i++)
#define repx(i, a, b) for(int i = a; i < b; i++)
LL f[5];

int main(){
    int L, ai;
    scanf("%d", &L);
    repx(i, 1, L + 1){
        scanf("%d", &ai);
        LL mn = f[0];
        f[0] += ai;
        mn = min(mn, f[1]), f[1] = mn + (ai ? ai & 1 : 2);
        mn = min(mn, f[2]), f[2] = mn + (ai ? (ai + 1) & 1 : 1);
        mn = min(mn, f[3]), f[3] = mn + (ai ? ai & 1 : 2);
        mn = min(mn, f[4]), f[4] = mn + ai;
    }
    printf("%lld\n", min(f[2], min(f[3], f[4])));
    return 0;
}
### Prompt

In cpp, your task is to solve the following problem:

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

### Response

```cpp
// ※※※ 解答不能 ※※※
// SovietPower氏.
// https://atcoder.jp/contests/yahoo-procon2019-qual/submissions/4241327
#include <bits/stdc++.h>
using namespace std;
using LL = long long;
#define rep(i, n) for(int i = 0; i < n; i++)
#define repx(i, a, b) for(int i = a; i < b; i++)
LL f[5];

int main(){
    int L, ai;
    scanf("%d", &L);
    repx(i, 1, L + 1){
        scanf("%d", &ai);
        LL mn = f[0];
        f[0] += ai;
        mn = min(mn, f[1]), f[1] = mn + (ai ? ai & 1 : 2);
        mn = min(mn, f[2]), f[2] = mn + (ai ? (ai + 1) & 1 : 1);
        mn = min(mn, f[3]), f[3] = mn + (ai ? ai & 1 : 2);
        mn = min(mn, f[4]), f[4] = mn + ai;
    }
    printf("%lld\n", min(f[2], min(f[3], f[4])));
    return 0;
}
```