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;

void chmin(int& a, int b) {
    a = min(a, b);
}

int main(void) {
    int L;cin>>L;
    int A[L];
    for (int i=0;i<L;i++) cin>>A[i];
    int dp[L+1][5];
    for(int i=0;i<=L;i++) for(int j=0;j<5;j++) dp[i][j]=1e9;
    dp[0][0] = 0;
    for(int i=0;i<L;i++) {
        int pre=dp[i][0];
        int diff;
        for(int j=0;j<5;j++){
            // 遷移元の最小を探す。この書き方だと、preのjは求めるj以下に必ずなる
            chmin(pre, dp[i][j]);
            if(j==0 || j==4) {
                diff = A[i];
            } else if(j==1 || j==3) {
                diff = (A[i]==0 ? 2 : A[i]%2);
            } else {
                diff = (A[i]+1)%2;
            }
            chmin(dp[i+1][j], pre+diff);
        }
    }
    int ans = dp[L][0];
    for(int j=1;j<5;j++) chmin(ans, dp[L][j]);
    cout << ans;
    return 0;
}