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"
using namespace std;
typedef long long ll;
#define INF (1<<30)
#define INFLL (1ll<<60)
typedef pair<int, int> P;
typedef pair<int, P> E;
#define MOD (1000000007ll)
#define l_ength size

void mul_mod(ll& a, ll b){
	a *= b;
	a %= MOD;
}

void add_mod(ll& a, ll b){
	b += MOD;
	a += b;
	a %= MOD;
}

int n;
ll a[225816],memo[2][225816];
bool done[2][225816];

ll f(int i){
	ll s,t,c = 0ll;
	if(done[0][i]){
		return memo[0][i];
	}
	done[0][i] = true;
	if(i==n-1){
		memo[0][i] = 0ll;
		return memo[0][i];
	}
	s = a[i];
	t = a[i+1];
	while(s > t){
		t *= 4ll;
		++c;
	}
	memo[0][i] = f(i+1)+c*(n-i-1);
	return memo[0][i];
}

ll g(int i){
	ll s,t,c = 0ll;
	if(done[1][i]){
		return memo[1][i];
	}
	done[1][i] = true;
	if(i==0){
		memo[1][i] = 0ll;
		return memo[1][i];
	}
	s = a[i];
	t = a[i-1];
	while(s > t){
		t *= 4ll;
		++c;
	}
	memo[1][i] = f(i-1)+c*(n-i-1);
	return memo[1][i];
}

int main(void){
	int i;
	ll ans;
	cin >> n;
	fill(done[0],done[2],false);
	for(i=0; i<n; ++i){
		cin >> a[i];
	}
	ans = min(2*f(0),n+2*g(n-1));
	for(i=1; i<n; ++i){
		ans = min(ans,i+2*f(i)+2*g(i-1));
	}
	cout << ans << endl;
	return 0;
}