Problem:
Score : 800 points

Problem Statement

There are N positive integers A_1, A_2, ..., A_N.
Takahashi can perform the following operation on these integers any number of times:

Choose 1 \leq i \leq N and multiply the value of A_i by -2.

Notice that he multiplies it by minus two.

He would like to make A_1 \leq A_2 \leq ... \leq A_N holds.
Find the minimum number of operations required. If it is impossible, print -1.

Constraints

1 \leq N \leq 200000

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the answer.

Sample Input 1

4
3 1 4 1

Sample Output 1

3

One possible solution is:

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now 3, 1, 4, -2.

Choose i=1 and multiply the value of A_1 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, -2.

Choose i=4 and multiply the value of A_4 by -2. A_1, A_2, A_3, A_4 are now -6, 1, 4, 4.

Sample Input 2

5
1 2 3 4 5

Sample Output 2

0

A_1 \leq A_2 \leq ... \leq A_N holds before any operation is performed.

Sample Input 3

8
657312726 129662684 181537270 324043958 468214806 916875077 825989291 319670097

Sample Output 3

7

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

#include <bits/stdc++.h>

using namespace std;
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
constexpr int MOD = 1000000007;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
constexpr int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};
constexpr int dy[] = {0, -1, 0, 1, 1, -1, -1, 1};

template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){os << "["; for (const auto &v : vec) {os << v << ","; } os << "]"; return os; }
template <typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &p){os << "(" << p.first << ", " << p.second << ")"; return os;}

void solve() {
    int N;
    cin >> N;
    vector<ll> A(N);
    for (int i = 0; i < N; i++) {
        cin >> A[i];
    }

    // i: マイナスとプラスの区切れ目
    ll ans = 1LL << 60;
    for(int i = 0; i <= N; i++) {
        vector<ll> B = A;
        ll cnt = 0;
        ll now = 0;
        for(int j = i - 1; j >= 0; j--) {
            while(now < B[j]) {
                B[j] *= -2;
                cnt++;
            }
            now = B[j];
        }
        for(int j = i; j < N; j++) {
            cout << j << endl;
            while(now > B[j]) {
                B[j] *= -2;
                cnt++;
            }
            now = B[j];
        }
        ans = min(ans, cnt);
    }
    cout << ans << endl;
}

int main() {
    std::cin.tie(0);
    std::ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    solve();
    return 0;
}