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 N (200009)
#define LL long long
using namespace std;

LL n,a[N],f[N][21],ans=1e18;

int main()
{
	memset(f,0x3f,sizeof(f));
	scanf("%lld",&n);
	for (int i=1; i<=n; ++i) scanf("%lld",&a[i]);
	for (int i=0; i<=20; ++i) f[1][i]=i;
	for (int i=2; i<=n; ++i)
	{
		for (int j=0; j<=20; ++j)
			for (int k=0; k<=20; ++k)
				if (a[i-1]*(1<<j)*((j%2==1)?-1:1) <=a[i]*(1<<k)*((k%2==1)?-1:1))
					f[i][k]=min(f[i][k],f[i-1][j]+k);
	}
	for (int i=0; i<=20; ++i)
		ans=min(ans,f[n][i]);
	cout<<ans<<endl;
}