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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, m, n) for(int(i) = (int)(m); i < (int)(n); ++i)
#define rep2(i, m, n) for(int(i) = (int)(n)-1; i >= (int)(m); --i)
#define REP(i, n) rep(i, 0, n)
#define REP2(i, n) rep2(i, 0, n)
#define all(hoge) (hoge).begin(), (hoge).end()
#define en '\n'
using ll = long long;
using ull = unsigned long long;
template <class T>
using vec = vector<T>;
template <class T>
using vvec = vector<vec<T>>;
typedef pair<ll, ll> P;
constexpr long long INF = 1LL << 60;
constexpr int INF_INT = 1 << 25;
constexpr long long MOD = (ll)1e9 + 7;
// constexpr long long MOD = 998244353LL;
using ld = long double;
static const ld pi = 3.141592653589793L;
typedef vector<ll> Array;
typedef vector<Array> Matrix;

template <class T>
inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

struct Edge {
    ll to, rev;
    long double cap;
    Edge(ll _to, long double _cap, ll _rev) {
        to = _to;
        cap = _cap;
        rev = _rev;
    }
};
using Edges = vector<Edge>;
using Graph = vector<Edges>;

void add_edge(Graph &G, ll from, ll to, long double cap, bool revFlag,
              long double revCap) {
    G[from].push_back(Edge(to, cap, (ll)G[to].size()));
    if(revFlag)
        G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));
}

template <int mod>
struct ModInt {
    int x;

    ModInt() : x(0) {}

    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

    ModInt &operator+=(const ModInt &p) {
        if((x += p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator-=(const ModInt &p) {
        if((x += mod - p.x) >= mod)
            x -= mod;
        return *this;
    }

    ModInt &operator*=(const ModInt &p) {
        x = (int)(1LL * x * p.x % mod);
        return *this;
    }

    ModInt &operator/=(const ModInt &p) {
        *this *= p.inverse();
        return *this;
    }

    ModInt operator-() const { return ModInt(-x); }

    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

    bool operator==(const ModInt &p) const { return x == p.x; }

    bool operator!=(const ModInt &p) const { return x != p.x; }

    ModInt inverse() const {
        int a = x, b = mod, u = 1, v = 0, t;
        while(b > 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
        }
        return ModInt(u);
    }

    ModInt pow(int64_t n) const {
        ModInt ret(1), mul(x);
        while(n > 0) {
            if(n & 1)
                ret *= mul;
            mul *= mul;
            n >>= 1;
        }
        return ret;
    }

    friend ostream &operator<<(ostream &os, const ModInt &p) {
        return os << p.x;
    }

    friend istream &operator>>(istream &is, ModInt &a) {
        int64_t t;
        is >> t;
        a = ModInt<mod>(t);
        return (is);
    }

    static int get_mod() { return mod; }
};

using mint = ModInt<MOD>;

void solve() {
    ll n;
    cin >> n;
    vec<ll> a(n);
    REP(i, n) { cin >> a[i]; }

    vvec<ll> dp(n + 1, vec<ll>(5, INF));
    dp[0][0] = 0;
    REP(i, n) {
        chmin(dp[i + 1][0], dp[i][0] + a[i]);

        chmin(dp[i + 1][1], dp[i][0] + (a[i] == 0 ? 2 : a[i] % 2));
        chmin(dp[i + 1][1], dp[i][1] + (a[i] == 0 ? 2 : a[i] % 2));

        chmin(dp[i + 1][2], dp[i][0] + (1 - a[i] % 2));
        chmin(dp[i + 1][2], dp[i][1] + (1 - a[i] % 2));
        chmin(dp[i + 1][2], dp[i][2] + (1 - a[i] % 2));

        chmin(dp[i + 1][3], dp[i][0] + (a[i] == 0 ? 2 : a[i] % 2));
        chmin(dp[i + 1][3], dp[i][1] + (a[i] == 0 ? 2 : a[i] % 2));
        chmin(dp[i + 1][3], dp[i][2] + (a[i] == 0 ? 2 : a[i] % 2));
        chmin(dp[i + 1][3], dp[i][3] + (a[i] == 0 ? 2 : a[i] % 2));

        chmin(dp[i + 1][4], dp[i][0] + a[i]);
        chmin(dp[i + 1][4], dp[i][1] + a[i]);
        chmin(dp[i + 1][4], dp[i][2] + a[i]);
        chmin(dp[i + 1][4], dp[i][3] + a[i]);
        chmin(dp[i + 1][4], dp[i][4] + a[i]);
    }

    ll ans = INF;
    REP(j, 5) { chmin(ans, dp[n][j]); }
    cout << ans << en;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    solve();
    // ll t;cin>>t;REP(i,t) solve();

    return 0;
}