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;
typedef long long ll;

const ll oo = 1e18;
const int N = 200005;

int n;
int v[N];

// 0, even, odd, even, 0
// 0, 1,     2,   3,  4

ll dp[N][5];
ll f(int i, int j){
    if(i == n) return 0;
    ll &ans = dp[i][j];
    if(~ans) return ans;
    ans = 1e18;
    if(j == 0) ans = f(i + 1, j) + v[i];
    if(j == 1) ans = f(i + 1, j) + (v[i] == 0 ? 2 : v[i] % 2);
    if(j == 2) ans = f(i + 1, j) + !(v[i] % 2);
    if(j == 3) ans = f(i + 1, j) + (v[i] == 0 ? 2 : v[i] % 2);
    if(j == 4) ans = f(i + 1, j) + v[i];
    if(j < 4) ans = min(ans, f(i, j + 1));
    return ans;
}

int main(){
    memset(dp, -1, sizeof dp);

    scanf("%d", &n);

    for(int i = 0; i < n; i++)
        scanf("%d", v + i);

    printf("%lld\n", f(0, 0));
}
