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: Runtime Error)

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

long long N, X[100009], E[100009], G;

vector<long long> solve() {
	G = X[0];
	for (int i = 1; i < N; i++) {
		if (G <= X[i]) { G = X[i]; E[i] = 0; }
		else {
			if (X[i - 1] >= X[i]) {
				for (int j = 0; j < 32; j++) {
					if (X[i - 1] <= (1LL << j)*X[i]) { E[i] = E[i - 1] + j; break; }
				}
			}
			else {
				for (int j = 0; j < 32; j++) {
					if ((1LL << j) * X[i - 1] > X[i]) { E[i] = E[i - 1] - (j - 1); break; }
				}
			}
		}
	}

	long long sum = 0, ret = 0, pos = 0; for (int i = 0; i < N; i++) sum += E[i];
	vector<long long>res; res.push_back(sum);

	for (int i = 0; i < N - 1; i++) {
		for (int j = i + 1; j <= N; j++) {
			long long JURAI = E[j]; if (j > pos) JURAI -= ret;

			if (j == i + 1) {
				E[i + 1] = 0;
			}
			else {
				if (G <= X[j]) { G = X[j]; E[j] = 0; }
				else {
					if (X[j - 1] >= X[j]) {
						for (int k = 0; k < 32; k++) {
							if (X[j - 1] <= (1LL << k)*X[j]) { E[j] = E[j - 1] + k; break; }
						}
					}
					else {
						for (int k = 0; k < 32; k++) {
							if ((1LL << k) * X[j - 1] > X[j]) { E[j] = E[j - 1] - (k - 1); break; }
						}
					}
				}
			}
			sum -= (JURAI - E[j]);
			if (E[j] >= 30 || j == N - 1 || JURAI - E[j] == 0) {
				ret = JURAI - E[j]; pos = j;
				res.push_back(sum - ret * (N - pos - 1));
			}
		}
	}
	res.push_back(0);
	return res;
}

int main() {
	cin >> N;
	for (int i = 0; i < N; i++) cin >> X[i];

	vector<long long> v1 = solve(); reverse(X, X + N);
	vector<long long> v2 = solve();
	reverse(v2.begin(), v2.end());
	long long minx = (1LL << 60);
	for (int i = 0; i <= N; i++) minx = min(minx, v1[i] + v2[i] + i);
	cout << minx << endl;
	return 0;
}