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 <iostream>
#include <string>
#include <cstring>
#include <vector>
#include <algorithm>
#include <set>
#include <stack>
#include <deque>
#include <cmath>
#include <map>
using namespace std;
using ll = long long;
#define rep(i, n) for (ll i = 0; i < n; i++)
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

int main() {
    ll n;
    cin >> n;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];
    vector<ll> dp(5);
    rep(i, n) {
        dp[0] += a[i]; // 0

        if (a[i] == 0) dp[1] += 2; // even > 0
        else dp[1] += a[i] % 2;

        dp[2] += (a[i] + 1) % 2; // odd > 0

        if (a[i] == 0) dp[3] += 2; // even > 0
        else dp[3] += a[i] % 2;

        dp[4] += a[i]; // 0

        rep(k, 4) chmin(dp[k+1], dp[k]);
    }
    cout << dp[4] << endl;
}