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: Accepted)

#include <bits/stdc++.h>
      
#define FOR(i,a,b) for(ll i = (a); i < (ll)(b); i++)
#define REP(i,n) FOR(i,0,n)
#define YYS(x,arr) for(auto& x:arr)
#define PW(x) (1LL<<(x))
#define SZ(x) ((ll)(x).size())

#define pb emplace_back
#define fi first
#define se second

using namespace std;

using ld = long double;
using ll = long long int;

const ll INF = (ll)1e9 + 10;
const ll INFLL = (ll)1e18 + 10;
const ll MOD = 1e9+7;
     
template<class T> T &chmin( T &a , const T &b ){ return a = min(a,b); }
template<class T> T &chmax( T &a , const T &b ){ return a = max(a,b); }
template<class T> void UNIQUE(vector<T> &a){ a.erase(unique(a.begin(), a.end()), a.end()); }

template<class S, class T> ostream& operator << (ostream& os, const pair<S, T> v){
  os << "(" << v.first << ", " << v.second << ")"; return os;
}
template<class T> ostream& operator << (ostream& os, const vector<T> v){
  for(int i = 0; i < v.size(); i++){if(i > 0){os << " ";} os << v[i];} return os;
}
template<class T> ostream& operator << (ostream& os, const vector<vector<T>> v){
  for(int i = 0; i < v.size(); i++){if(i > 0){os << endl;} os << v[i];} return os;
}

ll in(){long long int x; scanf("%lld", &x); return x;}
ld fin(){double x; scanf("%lf", &x); return x;}
char yuyushiki[1000010]; string stin(){scanf( "%s", yuyushiki ); return yuyushiki;}

// head
using T = tuple<ll, ll, ll>;

vector<ll> get(vector<ll> a){
  ll n = SZ(a);
  vector<ll> res(0);
  stack<T> st;
  ll cnt = 0;
  res.pb(0);
  REP(i, n){
    ll last = a[i];
    ll csz = 1;
    while(!st.empty()){
      ll x, y, sz;
      tie(x, y, sz) = st.top();
      if(y >= last){
        break;
      }
      st.pop();
      while(y < last){
        x *= 4;
        y *= 4;
        cnt += sz;
      }
      csz += sz;
      last = x;
    }
    st.emplace(last, a[i], csz);
    res.pb(cnt);
  }
  return res;
}

int main(){

  ll n = in();
  vector<ll> a(n);
  REP(i, n){
    a[i] = in();
  }

  reverse(a.begin(), a.end());
  vector<ll> f = get(a);
  REP(i, n){
    a[i] = a[i] * 2;
  }
  reverse(a.begin(), a.end());
  vector<ll> g = get(a);

  // cout << a << endl;
  // cout << f << endl;
  // cout << g << endl;

  ll ans = INFLL;
  REP(i, n+1){
    chmin(ans, (f[i] + g[n-i]) * 2 + n-i);
  }

  cout << ans << endl;
  
  return 0;
}