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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#line 2 "/Users/kaage/Desktop/ProgrammingWorkspace/library/other/template.hpp"
#define _CRT_SECURE_NO_WARNINGS
#pragma target("avx2")
#pragma optimize("O3")
#pragma optimize("unroll-loops")
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cfloat>
#include <climits>
#include <cmath>
#include <complex>
#include <ctime>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <memory>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <string.h>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define all(V) V.begin(),V.end()
typedef unsigned int uint;
typedef long long lint;
typedef unsigned long long ulint;
typedef std::pair<int, int> P;
typedef std::pair<lint, lint> LP;
constexpr int INF = INT_MAX/2;
constexpr lint LINF = LLONG_MAX/2;
constexpr double eps = DBL_EPSILON;
constexpr double PI=acos(-1);
template<class T>
class prique :public std::priority_queue<T, std::vector<T>, std::greater<T>> {};
template <class T, class U>
inline bool chmax(T& lhs, const U& rhs) {
	if (lhs < rhs) {
		lhs = rhs;
		return 1;
	}
	return 0;
}
template <class T, class U>
inline bool chmin(T& lhs, const U& rhs) {
	if (lhs > rhs) {
		lhs = rhs;
		return 1;
	}
	return 0;
}
inline lint gcd(lint a, lint b) {
	while (b) {
		lint c = a;
		a = b; b = c % b;
	}
	return a;
}
inline lint lcm(lint a, lint b) {
	return a / gcd(a, b) * b;
}
bool isprime(lint n) {
	if (n == 1)return false;
	for (int i = 2; i * i <= n; i++) {
		if (n % i == 0)return false;
	}
	return true;
}
template<typename T>
T mypow(T a, lint b) {
	T res(1);
	while(b){
		if(b&1)res*=a;
		a*=a;
		b>>=1;
	}
	return res;
}
lint modpow(lint a, lint b, lint m) {
	lint res(1);
	while(b){
		if(b&1){
			res*=a;res%=m;
		}
		a*=a;a%=m;
		b>>=1;
	}
	return res;
}
template<typename T>
void printArray(std::vector<T>& vec) {
	rep(i, vec.size()){
		std::cout << vec[i];
		std::cout<<(i==(int)vec.size()-1?"\n":" ");
	}
}
template<typename T>
void printArray(T l, T r) {
	T rprev = std::prev(r);
	for (T i = l; i != rprev; i++) {
		std::cout << *i << " ";
	}
	std::cout << *rprev << std::endl;
}
LP extGcd(lint a,lint b) {
	if(b==0)return {1,0};
	LP s=extGcd(b,a%b);
	std::swap(s.first,s.second);
	s.second-=a/b*s.first;
	return s;
}
LP ChineseRem(const lint& b1,const lint& m1,const lint& b2,const lint& m2) {
	lint p=extGcd(m1,m2).first;
	lint tmp=(b2-b1)*p%m2;
	lint r=(b1+m1*tmp+m1*m2)%(m1*m2);
	return std::make_pair(r,m1*m2);
}
template<typename F>
inline constexpr decltype(auto) lambda_fix(F&& f){
	return [f=std::forward<F>(f)](auto&&... args){
		return f(f,std::forward<decltype(args)>(args)...);
	};
}
#line 2 "main.cpp"
int N;
lint A[200010],B[200010],L[200010],R[200010];
int main(){
	std::cin>>N;
	rep(i,N)std::cin>>A[i];
	std::copy(A,A+N,B);
	L[0]=1;
	B[0]*=-2;
	std::vector<P> vec;
	REP(i,N-1){
		L[i]=L[i-1]+1;
		B[i]*=-2;
		if(B[i-1]<B[i]){
			lint r=B[i-1]/B[i];
			int cnt=0;
			while(r>=4){
				r/=4;
				cnt++;
			}
			if(cnt)vec.emplace_back(i,cnt);
		}
		else{
			lint r=(B[i]+B[i-1]-1)/B[i-1];
			int cnt=0;
			while(r>1){
				r/=4;
				cnt++;
			}
			rep(j,cnt){
				if(vec.empty())L[i]+=i*2;
				else{
					L[i]+=(i-vec.back().first)*2;
					if(--vec.back().second==0)vec.pop_back();
				}
			}
		}
	}
	vec.clear();
	std::copy(A,A+N,B);
	for(int i=N-2;i>=0;i--){
		R[i]=R[i+1];
		if(B[i]<B[i+1]){
			lint r=B[i+1]/B[i];
			int cnt=0;
			while(r>=4){
				r/=4;
				cnt++;
			}
			if(cnt)vec.emplace_back(i,cnt);
		}
		else{
			lint r=(B[i]+B[i+1]-1)/B[i+1];
			int cnt=0;
			while(r>1){
				r/=4;
				cnt++;
			}
			rep(j,cnt){
				if(vec.empty())R[i]+=(N-i-1)*2;
				else{
					R[i]+=(vec.back().first-i)*2;
					if(--vec.back().second==0)vec.pop_back();
				}
			}
		}
	}
	//printArray(L,L+N);
	//printArray(R,R+N);
	lint ans=std::min(L[N-1],R[0]);
	rep(i,N-1)chmin(ans,L[i]+R[i+1]);
	std::cout<<ans<<std::endl;
	return 0;
}