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 <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>

#include <iostream>
#include <complex>
#include <string>
#include <algorithm>
#include <numeric>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>

#include <functional>
#include <cassert>

typedef long long ll;
using namespace std;

#ifdef LOCAL
#define debug(x) cerr << __LINE__ << " : " << #x << " = " << (x) << endl;
#else
#define debug(x) ;
#endif

#define mod 1000000007 //1e9+7(prime number)
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define SIZE 200010


int log4(ll p, ll q, bool f = true){
  int counter = 0;
  int cc = !!(p % q) * 2;
  ll v = p / q;
  while(v >= 4){
    if(v % 4) cc++;
    v /= 4;
    counter++;
  }
  if(v > 1) cc++; 
  debug(p); debug(q); debug(cc);
  return counter + (cc >= 1) * f;
}

ll pow2(int x) {
  ll res = 1;
  for(int i=0;i<x;i++) res *= 2;
  return res;
}

int main(){
  int n;
  ll a[SIZE], dp1[SIZE], dp2[SIZE];

  scanf("%d", &n);

  for(int i=0;i<n;i++){
    scanf("%lld", a+i);
  }

  ll calc1 = 0, calc2 = 0;

  stack<pair<ll,int> > st, st2;
  st.push({LLINF, 0});
  st2.push({LLINF, n-1});

  dp1[0] = 1;
  ll prev = a[0];
  for(int i=1;i<n;i++){
    ll now = a[i];

    if(prev < now){
      ll r = log4(now, prev) * 2;
      prev *= pow2(r);
      while(r){
        auto p = st.top(); st.pop();
        ll r2 = min(p.first, r);
        r -= r2;
        p.first -= r2;
        calc1 += (ll)r2 * p.second;
        if(p.first > 0) st.push(p);
      }
    }

    ll e = log4(prev, now, false) * 2;
    
    if(e){
      st.push({e, i});
    }

    dp1[i] = calc1 + i + 1;
    debug(dp1[i]);
    prev = now;
  }


  dp2[n] = dp2[n-1] = 0;
  prev = a[n-1];
  for(int i=n-2;i>=0;i--){
    ll now = a[i];

    if(prev < now){
      ll r = log4(now, prev) * 2;
      prev *= pow2(r);
      while(r){
        auto p = st2.top(); st2.pop();
        ll r2 = min(p.first, r);
        r -= r2;
        p.first -= r2;
        calc2 += (ll)r2 * (p.second - i);
        if(p.first > 0) st2.push(p);
      }
    }

    ll e = log4(prev, now, false) * 2;
    
    if(e){
      st2.push({e, i});
    }

    dp2[i] = calc2;
    prev = now;
  }


  ll ans = dp2[0];
  
  for(int i=0;i<n;i++){
    ans = min(ans, dp1[i] + dp2[i+1]);
    //cerr << dp1[i] << " + " << dp2[i+1] << endl;
  }

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