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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<algorithm>
#include<vector>
#include<stack>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int MAXN = 2e5 + 5;
const int inf = 0x3f3f3f3f;
const ll linf = 0x3f3f3f3f3f3f3f3f;

inline int calc(int x,int y)
{
	if(x>y)
	{
		int res = 0;
		while(x>y)
			++res, y<<=1;
		return -(res + (res&1));
	}
	else
	{
		int res = 0;
		while(x<y)
			++res, x<<=1;
		return res + (res&1);
	}
}

int a[MAXN];
int p[MAXN], q[MAXN];
ll pre_p[MAXN], suf_q[MAXN];

int main(void)
{
	int n;
	scanf("%d",&n);
	for(int i=1; i<=n; ++i) scanf("%d",&a[i]);
	
	stack<pii> sta;
	sta.push(make_pair(1,inf));
	for(int i=1; i<n; ++i)
	{
		p[i] = calc(a[i], a[i+1]);
		pre_p[i] = pre_p[i-1];
		
		if(p[i] < 0){ sta.push(make_pair(i, -p[i])); continue;}
		
		while(p[i])
		{
			int cur = min(p[i], sta.top().second);
			pre_p[i] += cur * (i - sta.top().first + 1);
			
			p[i] -= cur;
			sta.top().second -= cur;
			if(!sta.top().second) sta.pop();
		}
	}
	
	while(!sta.empty()) sta.pop();
	sta.push(make_pair(n,inf));
	for(int i=n; i>=2; --i)
	{
		q[i] = calc(a[i], a[i-1]);
		suf_q[i] = suf_q[i+1];
		
		if(q[i] < 0){ sta.push(make_pair(i, -q[i])); continue;}
		
		while(q[i])
		{
			int cur = min(q[i], sta.top().second);
			suf_q[i] += cur * (sta.top().first - i + 1);
			
			q[i] -= cur;
			sta.top().second -= cur;
			if(!sta.top().second) sta.pop();
		}
	}
	
	ll ans = linf;
	for(int i=0; i<=n; ++i)
	{
		ll res = 0;
		if(i>=1) res += pre_p[i-1] + i;
		if(i+1<=n) res += suf_q[i+2];
		
		ans = min(ans, res);
	}
	printf("%lld",ans);
	return 0;
}