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 ADD(a, b) a = (a + ll(b)) % mod
#define MUL(a, b) a = (a * ll(b)) % mod
#define MAX(a, b) a = max(a, b)
#define MIN(a, b) a = min(a, b)
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)
#define all(a) (a).begin(), (a).end()
#define sz(v) (int)(v).size()
#define pb push_back
#define sec second
#define fst first
#define debug(fmt, ...) Debug(__LINE__, ":", fmt, ##__VA_ARGS__)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<int, pi> ppi;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> mat;
typedef complex<double> comp;
void Debug() {cout << '\n'; }
template<class FIRST, class... REST>void Debug(FIRST arg, REST... rest){
	cout<<arg<<" ";Debug(rest...);}
template<class T>ostream& operator<<(ostream& out,const vector<T>& v) {
	out<<"[";if(!v.empty()){rep(i,0,sz(v)-1)out<<v[i]<<", ";out<<v.back();}out<<"]";return out;}
template<class S, class T>ostream& operator<<(ostream& out,const pair<S, T>& v){
	out<<"("<<v.first<<", "<<v.second<<")";return out;}
const int MAX_N = 300010;
const int MAX_V = 100010;
const double eps = 1e-6;
const ll mod = 1000000007;
const int inf = 1 << 30;
const ll linf = 1LL << 60;
const double PI = 3.14159265358979323846;
///////////////////////////////////////////////////////////////////////////////////////////////////

int N;
ll A[MAX_N];
ll B1[MAX_N], B2[MAX_N];

void pre(ll B[MAX_N]) {
	stack<pi> st;
	st.push(pl(linf, N - 1));
	B[N - 1] = 0;
	ll cost = 0;
	rer(i, N - 1, 0) {
		// debug(i);
		while(A[i] > A[i + 1]) {
			A[i + 1] *= 4;
			pl p = st.top(); st.pop();
			p.fst--;
			cost += (2 * (p.sec - i));
			if(p.fst) st.push(p);
		}
		B[i] = cost;
		ll n = A[i];
		int cnt = 0;
		while(n * 4 <= A[i + 1]) {
			n *= 4; cnt++;
		}
		if(cnt) st.push(pi(cnt, i));
	}
}

void solve() {
	cin >> N;
	rep(i, 0, N) cin >> A[i];
	pre(B1);
	reverse(A, A + N);
	pre(B2);
	reverse(B2, B2 + N);
	// debug(vi(B1, B1 + N));
	// debug(vi(B2, B2 + N));
	ll res = B1[0];
	MIN(res, B2[N - 1] + N);
	rep(i, 0, N - 1) {
		MIN(res, B2[i] + B1[i + 1] + i + 1);
	}
	cout << res << "\n";
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
    cin.tie(0);
#endif
    cout << fixed;
	cout.precision(20);
	srand((unsigned int)time(NULL));
#ifdef LOCAL
	//freopen("in.txt", "wt", stdout); //for tester
    freopen("in.txt", "rt", stdin);
#endif	
	solve();
#ifdef LOCAL
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
#endif
	return 0;
}