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: Accepted)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
#define pb push_back
#define fi first
#define se second
#define Mod 1000000007


int main(){
    ll i,j,k;
    ll n;
    cin>>n;
    long double a[n],b[n];
    for(i=0;i<n;i++){
        ll x;
        cin>>x;
        a[i]=x;
        b[i]=x;
    }

    ll r[n],l[n];
    r[n-1]=0;
    ll r1[n],l1[n];

    stack<ll> s1;

    for(i=n-2;i>=0;i--){
        r[i]=r[i+1];
        if(a[i]<a[i+1]){
            ll w=a[i],w1=a[i+1];
            while(w<=w1){
                w*=4;
                if(w<=w1) s1.push(i);
            }
        }

        while (a[i]>a[i+1]){
            if(!s1.empty()) {
                r[i]+=2*(s1.top()-i);
                s1.pop();
            }
            else r[i]+=2*(n-i-1);
            a[i+1]*=4;
        }
    }
        
    
    stack<ll> s2;

    
    l[0]=1;
    for(i=1;i<n;i++){
        l[i]=l[i-1]+1;
        ll w1=b[i],w2=b[i-1];
        if(w1<=w2){
            while(w1<=w2){
            w1*=4;
            if(w1<=w2) s2.push(i);
            }
        }
        else {
        while (w1>w2){
            if(!s2.empty()) {
                l[i]+=2*(i-s2.top());
                s2.pop();
            }
            else l[i]+=2*i;
            w2*=4;
        }
        }
    }

    ll ans=1e18;
    for(i=1;i<n;i++){
        if(ans>l[i-1]+r[i]) ans=l[i-1]+r[i];
    }
    if(ans>l[n-1]) ans=l[n-1];
    if(ans>r[0]) ans=r[0];
    cout<<ans<<endl;
    return 0;
}