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>

#define int long long
using namespace std;
#define rep(i, n) for(int i=0;i<(n);++i)
typedef pair<int, int> pii;
const int INF = 1l << 60;
#define u_b upper_bound
#define l_b lower_bound

int N;
int A[200200];
int X[200200], Y[200200];

void preCalc() {
    int prev = 0;
    for (int i = 1; i <= N; ++i) {
        int cnt = 1;
        int a = A[i - 1];
        if (prev == 0) {
            prev = a;
            X[i] = 1;
            continue;
        }
        while (a > prev) {
            prev *= 4;
            cnt += 2 * (i - 1);
        }
        prev = a;
        X[i] = cnt + X[i - 1];
    }

    prev = 0;
    for (int i = N - 1; i >= 0; --i) {
        int cnt = 0;
        int a = A[i];
        if (prev == 0) {
            prev = a;
            continue;
        }
        while (a > prev) {
            prev *= 4;
            cnt += 2 * (N - i - 1);
        }
        prev = a;
        Y[i] = cnt + Y[i + 1];
    }
}

signed main() {
    cin >> N;
    rep(i, N) {
        cin >> A[i];
    }
    int ans = INF;
    preCalc();
    for (int i = 0; i <= N; ++i) {
        //i-1番目まで負
        //i番目から正
        ans = min(ans, X[i] + Y[i]);
    }
    cout << ans << endl;
    return 0;
}