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<bits/stdc++.h>
#define st first
#define nd second
#define pb push_back
#define ppb pop_back
#define ppf pop_front
#define umax(x,y) x=max(x,y)
#define umin(x,y) x=min(x,y)
#define ll long long
#define ii pair<int,int>
#define iii pair<ii,int>
#define iiii pair<ii,ii>
#define sz(x) ((int) x.size())
#define orta ((bas+son)>>1)
#define all(x) x.begin(),x.end()
#define dbgs(x) cerr<<(#x)<<" --> "<<(x)<<" "
#define dbg(x) cerr<<(#x)<<" --> "<<(x)<<endl;getchar()
#define pw(x) (1<<(x))
#define inf 2000000000
#define MOD 2018012016
#define N 200005
#define M 1000005
#define LOG 1000000
#define KOK 32000
#define EPS 0.000000001
using namespace std;

int n;
ll ans,a[N],b[N],pre[N],suf[N];

int main() {

//	freopen("input.txt","r",stdin);

	scanf("%d",&n);

	for(int i=1;i<=n;i++) {

		scanf("%lld",a+i);

		b[i]=a[i];

	}

	a[1]*=2;

	pre[1]=1;

	for(int i=2;i<=n;i++) {

		ll cur=1;

		a[i]*=2;

		while(a[i-1]<a[i]) {

			a[i-1]*=4ll;

			cur+=2*(i-1);

		}

		pre[i]=pre[i-1]+cur;

	}

	suf[n]=0;

	for(int i=n-1;i>=1;i--) {

		ll cur=0;

		while(b[i+1]<b[i]) {

			b[i+1]*=4ll;

			cur+=2*(n-i);

		}

		suf[i]=suf[i+1]+cur;

	}

	ans=pre[0]+suf[1];

	for(int i=1;i<=n;i++) {

		umin(ans,pre[i]+suf[i+1]);

	}

	printf("%lld",ans);

}