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 <iostream>
#include <stdio.h>
#include <fstream>
#include <algorithm>
#include <string>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <limits.h>
#include <math.h>
#include <functional>
#include <bitset>
#include <iomanip>
#include <cassert>

#define repeat(i,n) for (long long i = 0; (i) < (n); ++ (i))
#define debug(x) cerr << #x << ": " << x << '\n'
#define debugArray(x,n) for(long long i = 0; (i) < (n); ++ (i)) cerr << #x << "[" << i << "]: " << x[i] << '\n'
#define debugArrayP(x,n) for(long long i = 0; (i) < (n); ++ (i)) cerr << #x << "[" << i << "]: " << x[i].first<< " " << x[i].second << '\n'

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> Pii;
typedef vector<int> vint;
typedef vector<ll> vll;
const ll INF = INT_MAX;
const ll MOD = 1e9+7;

int N;
ll A[212345];
ll dp[2][212345][16];

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin>>N;
  repeat(i,N){
    cin>>A[i];
  }
  for(int l=N-1;l>=0;l--){
    //debug(l);
    repeat(x,16){
      if(l==N-1){
        dp[0][l][x] = x;
      }else{
        ll cnt=0;
        ll a=1;
        while(A[l]>A[l+1]*a){
          a<<=2;
          cnt++;
        }
        a=1;
        ll c=0;
        while(A[l+1]>=A[l]*a){
          a<<=2;
          c++;
        }
        c = max(0ll,c-1);
        cnt += max(0ll,x-c);
        //debug(x);
        //debug(cnt);
        if(cnt<=15){
          dp[0][l][x] = x+dp[0][l+1][cnt];
        }else{
          dp[0][l][x] = x+dp[0][l+1][15]+(N-l-1)*(cnt-15);
        }
      }
    }
  }

  for(int l=1;l<N;l++){
    repeat(x,16){
      if(l==1){
        dp[1][l][x] = x;
      }else{
        ll cnt=0;
        ll a=1;
        while(A[l-1]>A[l-2]*a){
          a<<=2;
          cnt++;
        }
        a=1;
        ll c=0;
        while(A[l-2]>=A[l-1]*a){
          a<<=2;
          c++;
        }
        c = max(0ll,c-1);
        cnt += max(0ll,x-c);
        if(cnt<=15){
          dp[1][l][x] = x+dp[1][l-1][cnt];
        }else{
          dp[1][l][x] = x+dp[1][l-1][15]+(l-1)*(cnt-15);
        }
      }
    }
  }

  //debugArray(dp[0][3],15);
  ll ans=INF;
  for(int i=0;i<N;i++){
    ans = min(ans,i+2*dp[1][i][0]+2*dp[0][i][0]);
  }
  cout<<ans<<endl;
  return 0;
}