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>

typedef long long int ll;
typedef unsigned long long int ull;
#define rep(i,n) for(auto i=0;i<(n);i++)
#define rep1(i,n) for(auto i=1;i<=(n);i++)
#define all(x) (x).begin(),(x).end()

using namespace std;


struct init{ init(){ cin.tie(0); ios::sync_with_stdio(0); cout<<fixed<<setprecision(20); }; }aaaaaaa;


int main(){
    ll l;
    cin >> l;
    vector<ll> a(l+1, 0);
    rep(i, l) cin >> a[i+1];
    
    vector<vector<ll>> dp(l+1, vector<ll>(5, 0));
    rep(i, l+1){
        if(i == 0) continue;
        ll x = a[i];
        dp[i][0] = dp[i-1][0] + x;
        dp[i][1] = min(dp[i][0], dp[i-1][1] + (x? x%2:2));
        dp[i][2] = min(dp[i][1], dp[i-1][2]+((x+1)%2));
        dp[i][3] = min(dp[i][2], dp[i-1][3]+(x? x%2:2));
        dp[i][4] = min(dp[i][3], dp[i-1][4]+x);
    }

    cout << dp[l][4] << endl;

    // even odd even

    return 0;
}

