Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

4
1
0
2
3

Sample Output 1

1

Assume that Snuke walks as follows:

He starts walking at coordinate 3 and finishes walking at coordinate 4, visiting coordinates 3,4,3,2,3,4 in this order.

Then, Snuke's four ears will contain 0,0,2,3 stones, respectively.
Ringo can satisfy the requirement by putting one stone in the first ear.

Sample Input 2

8
2
0
0
2
1
3
4
1

Sample Output 2

3

Sample Input 3

7
314159265
358979323
846264338
327950288
419716939
937510582
0

Sample Output 3

1

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// template version 1.6
// {{{ define basic macro
using namespace std;
#include <iostream>
#include <bits/stdc++.h>
// varibable settings
#define int long long
#define INF 1000000000000000000LL
// #define INF 2147483647
#define MOD 1000000007LL
// #define MOD 998244353LL
#define infile "../test/sample-1.in"

#define _overload3(_1,_2,_3,name,...) name
#define _rep(i,n) repi(i,0,n)
#define repi(i,a,b) for(int i=(int)(a);i<(int)(b);++i)
#define rep(...) _overload3(__VA_ARGS__,repi,_rep,)(__VA_ARGS__)
#define _rrep(i,n) rrepi(i,0,n)
#define rrepi(i,a,b) for(int i=(int)(b-1);i>=(int)(a);--i)
#define rrep(...) _overload3(__VA_ARGS__,rrepi,_rrep,)(__VA_ARGS__)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define uni(x) sort(all(x));x.erase(unique(all(x)),x.end())
#define ten(n) ((int)1e##n)
template<class T, size_t N> size_t sza(const T (&array)[N]) { return N; }
template<class T> inline void chmax(T &a, const T &b) { if(a < b) a = b; }
template<class T> inline void chmin(T &a, const T &b) { if(a > b) a = b; }
template<class T=int> T in() {T x; cin>>x; return (x);}
struct Fast { Fast(){ std::cin.tie(0); ios::sync_with_stdio(false); } } fast;

// dump macro
#ifdef PCM
  #include "dump.hpp"
#else
  #define dump(...) 42
  #define dump_1d(...) 42
  #define dump_2d(...) 42
#endif

typedef long long ll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;
typedef pair<int,int> pii;
typedef tuple<int,int,int> iii;
template<typename T> using PQ = priority_queue<T, vector<T>, greater<T>>;
int dx[]={1, -1, 0, 0};
int dy[]={0, 0, 1, -1};
#define fi first
#define se second

// }}}

int solve(){
  int L;cin>>L;
  int a[L]; rep(i, L) {cin>>a[i];}

  int M = accumulate(a, a+L, 0LL);

  int dp[L]={};  // dp[i]: 区間iの右側にいる時のscoreの最大
  dp[0] = a[0];

  int mscore = 0;
  rep(i, 1, L){
    if (a[i]==0) {
      chmax(dp[i], dp[i-1]-1);
    }
    else if(a[i]%2==1){
      chmax(dp[i], dp[i-1]+a[i]);
      chmax(mscore, dp[i]);
    }
    else if(a[i]%2==0){
      chmax(dp[i], dp[i-1]+a[i]-1);
      chmax(dp[i], a[i]);  // new start

      chmax(mscore, dp[i]+1);
    }
  }

  dump_1d(dp, L);
  dump(M, mscore);

  cout << M-mscore << endl;

  return 0;
}

signed main() { //{{{
#ifdef INPUT_FROM_FILE
  std::ifstream in(infile);
  std::cin.rdbuf(in.rdbuf());
#endif
  solve();
  return 0;
} //}}}