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)

#include<bits/stdc++.h>
using namespace std;

#define ll long long int
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007

int main() 
{
    FIO;
    ll t,n,k,i,j,j1;
    cin >> n;
    
    ll a[n];
    pair<ll,ll> dp[n][31];
    for(i=0;i<n;i++)
        cin >> a[i];
    
    k=a[0];
    for(i=0;i<31;i++){
        dp[0][i]=make_pair(k,i);
        k*=-2;
    }
        
    for(i=1;i<n;i++){
        k=a[i];
        for(j=0;j<31;j++){
            dp[i][j]=make_pair(k,INT_MAX);
            k*=-2;
        }
    
        for(j1=0;j1<31;j1++){
            k=a[i];
            for(j=0;j<31;j++){
                
                if(dp[i][j].second>dp[i-1][j1].second+j && k>=dp[i-1][j1].first){
                    dp[i][j]=make_pair(k,dp[i-1][j1].second+j);
                }
                
                k*=-2;
            }
        }
    
    }
    
    k=LLONG_MAX;
    
    for(i=0;i<31;i++)
        k=min(k,dp[n-1][i].second);
    
    cout << k;
    
    
	return 0;
}