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;
using LL = long long;
using PII = pair<int, int>;
#define FI(i,a,b) for(int i=(a);i<=(b);++i)
#define FD(i,b,a) for(int i=(b);i>=(a);--i)
#define DEBUG(x) cerr << #x << ": " << (x) << endl;
constexpr LL inv(LL x, LL m) { return x > m ? inv(x % m, m) : x > 1 ? inv(m % x, m) * (m - m / x) % m : x; }
constexpr LL mpow(LL a, LL k, LL m) { LL r(1); for (a %= m; k; k >>= 1, a = a * a % m) if (k & 1)r = r * a % m; return r; }

const int N = 2e5 + 5;
LL n, A[N], R[N], L[N];

int main() {
    cin >> n;
    FI(i, 1, n) scanf("%d", A + i);
    FD(i, n - 1, 1) {
        if (A[i + 1] < A[i]) {
            LL x = A[i + 1];
            int cnt = 0;
            for (int j = 0; j < 32; j++) {
                x *= 2;
                cnt++;
                if (cnt & 1) continue;
                if (x >= A[i]) break;
            }
            R[i] = R[i + 1] + cnt * (n - i);
        } else R[i] = R[i + 1];
    }
    L[1] = 0, L[2] = 1;
    FI(i, 3, n) {
        if (A[i - 1] > A[i - 2]) {
            LL x = A[i - 2];
            int cnt = 0;
            for (int j = 0; j < 32; j++) {
                x *= 2;
                cnt++;
                if (cnt & 1) continue;
                if (x >= A[i - 1]) break;
            }
            assert(x >= A[i - 1]);
            L[i] = L[i - 1] + cnt * (i - 2) + 1;
        } else L[i] = L[i - 1] + 1;
    }
    LL ans = INT_MAX;
    FI(i, 1, n) ans = min(ans, L[i] + R[i]);
    cout << ans << endl;
}