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<algorithm>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
#define int ll
const ll inf=174417441744174417;
ll dp1[200005][20]={0},dp2[200005][20]={0},ar[200005]={0};
signed main(){
  ll n;
  cin>>n;
  for(ll i=1;i<=n;i++) cin>>ar[i];
  for(ll i=0;i<200005;i++){
    for(ll j=0;j<20;j++){
      dp1[i][j]=inf;
      dp2[i][j]=inf;
    }
  }
  for(ll i=0;i<=15;i++){
    dp1[n+1][i]=0;
    dp1[n][i]=i;
  }
  for(ll i=n-1;i>=1;i--){
    for(ll j=0;j<=15;j++){
      for(ll x=0;x<=15;x++){
        if((ar[i]*(1<<(2*j)))<=(ar[i+1]*(1<<(2*x)))){
          dp1[i][j]=min(dp1[i][j],dp1[i+1][x]+j);
        }
      }
    }
  }


  for(ll i=0;i<=15;i++) {
    dp2[0][i]=0;
    dp2[1][i]=i;
  }
  for(ll i=2;i<=n;i++){
    for(ll j=0;j<=15;j++){
      for(ll x=0;x<=15;x++){
        if(ar[i]*(1<<(2*j))<=ar[i-1]*(1<<(2*x))){
          dp2[i][j]=min(dp2[i][j],dp2[i-1][x]+j);
        }
      }
    }
  }
  ll ans=inf;
  for(ll i=0;i<=n;i++){
    for(ll j=0;j<=15;j++){
      for(ll k=0;k<=15;k++){
        ans=min(ans,i+(dp2[i][j]+dp1[i+1][k])*2);
      }
    }
  }
  cout<<ans<<endl;
}