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 <vector>
#include <cmath>
using namespace std;
typedef unsigned long long ll;
ll a[200010],f[200010] = {},g[200010] = {};
vector<int> v;
int main(){
	long long i,n;
	cin >> n;
	for(i=1;i<=n;i++){
		cin >> a[i];
		if(a[i]==0) v.push_back(i);
	}
	if(v.size()>=1){
		for(i=1;i<v.size();i++){
			if(v[i]-v[i-1]>1){
				cout << -1 << endl;
				return 0;
			}
		}
	}
	f[0] = 0;
	for(i=1;i<=n;i++){
		if(!a[i]) break;
		if(i==1){
			f[i] = 1;
			continue;
		}
		ll t = 0;
		while(a[i-1]*(ll)pow(4,t)<a[i]) t++;
		f[i] = f[i-1] + i*2*t + 1;
	}
	g[n+1] = 0;
	for(i=n;i>=1;i--){
		if(!a[i]) break;
		if(i==n){
			g[i] = 0;
			continue;
		}
		ll t = 0;
		while(a[i+1]*(ll)pow(4,t)<a[i]) t++;
		g[i] = g[i+1] + (n-i)*2*t;
	}
	/*for(i=0;i<=n;i++){
		cout << f[i] << " ";
	}
	cout << endl;
	for(i=1;i<=n;i++){
		cout << g[i] << " ";
	}
	cout << endl;*/
	if((int)v.size()){
		int x = v[0],y = v.back();
		cout << f[x-1] + g[y+1] << endl;
	}else{
		ll ans = 10000000000000000;
		for(i=0;i<=n;i++){
			ans = min(ans,f[i] + g[i+1]);
		}
		cout << ans << endl;
	}
}