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 main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int N;cin>>N;
  vll A(N);
  repeat(i,N){
    cin >> A[i];
  }
  cout << -1 << endl;
  return 0;
  ll ans = INF;
  ll high=N,low=-1;
  while(high-low>1){
    ll midlow = (high+low)/2;
    ll midhigh = (high+low)/2+1;
    //debug(midlow);
    ll alow = 0;
    ll precnt=0;
    for(int i=midlow-1;i>0;i--){
      ll x=1;
      ll cnt=0;
      while(A[i]>A[i-1]*x){
        x <<= 2;
        cnt += 2;
      }
      x = 1;
      ll c=0;
      while(A[i-1]>A[i]*x){
        x <<= 2;
        c+=2;
      }
      c=max(0ll,c-2);
      c=max(0ll,precnt-c);
      cnt+=c;
      alow += cnt;
      precnt=cnt;
    }
    alow += midlow;
    //debug(alow);
    precnt=0;
    for(int i=midlow;i<N-1;i++){
      //debug(i);
      ll x=1;
      ll cnt=0;
      while(A[i]>A[i+1]*x){
        x <<= 2;
        cnt += 2;
      }
      x = 1;
      ll c=0;
      while(A[i+1]>A[i]*x){
        x <<= 2;
        c+=2;
      }
      c=max(0ll,c-2);
      c=max(0ll,precnt-c);
      cnt+=c;
      //debug(cnt);
      alow += cnt;
      precnt=cnt;
    }
    //debug(alow);
    ll ahigh=0;
    precnt=0;
    for(int i=midhigh-1;i>0;i--){
      ll x=1;
      ll cnt=0;
      while(A[i]>A[i-1]*x){
        x <<= 2;
        cnt += 2;
      }
      x = 1;
      ll c=0;
      while(A[i-1]>A[i]*x){
        x <<= 2;
        c+=2;
      }
      c=max(0ll,c-2);
      c=max(0ll,precnt-c);
      cnt+=c;
      ahigh += cnt;
      precnt=cnt;
    }
    ahigh += midhigh;
    precnt=0;
    for(int i=midhigh;i<N-1;i++){
      ll x=1;
      ll cnt=0;
      while(A[i]>A[i+1]*x){
        x <<= 2;
        cnt += 2;
      }
      x = 1;
      ll c=0;
      while(A[i+1]>A[i]*x){
        x <<= 2;
        c+=2;
      }
      c=max(0ll,c-2);
      c=max(0ll,precnt-c);
      cnt+=c;
      ahigh += cnt;
      precnt=cnt;
    }
    if(alow<=ahigh){
      high = midlow;
      ans = min(ans,alow);
    }else{
      low = midlow;
      ans = min(ans,ahigh);
    }
  }
  cout << ans << endl;
  return 0;
}